Solved Examples

How to Use This Problem Bank

Eleven sections of theory, and now the part that actually earns marks. What follows is 50 worked problems, arranged easy first and hard last, covering the whole chapter — from "which formula is this?" up to the multi-step problems that decide ranks.

Work them with a pen and paper. Cover the solution, try it, then compare — including the check at the end of each one, because the check is where the marks usually leak away.

The four questions to ask before you write anything

Almost every mistake in thermal physics is made in the first ten seconds, before any arithmetic starts. Ask these four, in this order:

1. Does this expression need kelvin, or will Celsius do? If the temperature appears as a ratio, a product or a power — $\frac{T_2}{T_1}$, $T^4$, $PV = nRT$, $\lambda_m T = b$ — it must be in kelvin. If only a difference $\Delta T$ appears — $\Delta L = \alpha L\,\Delta T$, $Q = ms\,\Delta T$, $H = \frac{KA\Delta T}{L}$ — Celsius and kelvin give the identical number. Getting this wrong is the single commonest wrong answer in the chapter.
2. Can a phase change happen here? If ice, water and steam are all in play, you may not assume the ice all melts or the steam all condenses. Compute the heat available and the heat required, compare them, and only then decide.
3. Is this conduction, convection or radiation? A solid with two faces at different temperatures is conduction. A moving fluid is convection. A vacuum, a fourth power or a colour is radiation.
4. Series or parallel? Slabs stacked face to face share the same heat current and add their resistances. Slabs side by side share the same temperature difference and add their conductances. Deciding this first turns most conduction problems into one line.

Key Point — the master formulas, all in one place: $$t_F = \frac{9}{5}t_C + 32, \qquad T = t_C + 273.15, \qquad T = 273.16\,\frac{p}{p_{tr}}$$ $$\Delta L = \alpha L\,\Delta T, \qquad \Delta A = \beta A\,\Delta T, \qquad \Delta V = \gamma V\,\Delta T, \qquad \alpha : \beta : \gamma = 1 : 2 : 3$$ $$\gamma_{app} = \gamma_{real} - \gamma_{vessel}, \qquad \text{thermal stress} = Y\alpha\,\Delta T$$ $$Q = ms\,\Delta T, \qquad S = ms, \qquad C = \frac{1}{\mu}\frac{\Delta Q}{\Delta T}, \qquad Q = mL_f \ \text{or}\ mL_v$$ $$\sum \Delta Q = 0, \qquad W = \frac{m_c s_c}{s_w}$$ $$H = KA\frac{\Delta T}{L} = \frac{\Delta T}{R}, \qquad R = \frac{L}{KA}, \qquad R_{series} = R_1 + R_2, \qquad \frac{1}{R_{par}} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$\lambda_m T = b, \qquad H = \sigma A e\left(T^4 - T_s^4\right), \qquad -\frac{dT}{dt} = k(T - T_s)$$

A note on symbols, once

Throughout this chapter $T$ is an absolute temperature in kelvin and $t$ or $t_C$ is a Celsius temperature. A bare $L$ is a length; latent heat always carries a subscript, $L_f$ for fusion and $L_v$ for vaporisation. $s$ is specific heat capacity in J/(kg K), also written $c$, while $C$ is molar specific heat in J/(mol K) and $S$ is the heat capacity of a whole named body in J/K. $K$ is thermal conductivity, also written $k$ or $\lambda$; lowercase $k$ is reserved here for the cooling constant in Newton's law. In radiation, $a$ is absorptive power — also written $\alpha$ or $a_\lambda$, but here $\alpha$ belongs to linear expansion and nothing else — $e$ is emissivity, $\sigma$ is the Stefan-Boltzmann constant and $b$ is Wien's constant. $R$ is thermal resistance everywhere except in the molar-specific-heat problems, where it is the gas constant; each solution says which.

The data used below

Unless a problem states its own numbers, every solution here uses this table, and each problem also restates the constants it uses inside its own solution, so you never have to scroll back.

Quantity	Value used
Specific heat capacity of water	4186 J/(kg K)
Specific heat capacity of ice	2100 J/(kg K)
Specific heat capacity of steam	2010 J/(kg K)
Latent heat of fusion of ice, $L_f$	$3.33 \times 10^{5}$ J/kg
Latent heat of vaporisation of water, $L_v$	$22.6 \times 10^{5}$ J/kg
Coefficient of linear expansion $\alpha$: steel	$1.2 \times 10^{-5}$ K$^{-1}$
copper / brass / aluminium / glass	$1.7$ / $1.8$ / $2.3$ / $0.9$, each $\times 10^{-5}$ K$^{-1}$
Thermal conductivity $K$: copper / aluminium	385 / 205 W/(m K)
brass / steel / brick / glass	109 / 50.2 / 0.72 / 0.80 W/(m K)
water / thermacole (EPS) / still air	0.6 / 0.033 / 0.024 W/(m K)
Stefan-Boltzmann constant $\sigma$	$5.67 \times 10^{-8}$ W/(m$^2$ K$^4$)
Wien's displacement constant $b$	$2.9 \times 10^{-3}$ m K
Gas constant $R$	8.31 J/(mol K)
Triple point of water	273.16 K

On the value of $g$: the two problems that need it both use $9.8$ m/s$^2$, and say so. No problem here mixes $9.8$ and $10$.

[Board Important] Every solution below writes the formula on its own line before any number goes into it, and every temperature that enters a ratio or a fourth power is written with the letter K next to it. Do both in the exam. A correct formula with an arithmetic slip still earns most of the marks; a Celsius value inside a fourth power loses them all.

Solved Examples

Part 1: Temperature, Scales and Thermometers

Six warm-ups. Everything here is a straight line drawn between two known points, and the only real skill is deciding whether the problem wants an interval or a reading.

Key Point: A Celsius degree and a kelvin are the same size, so a temperature rise of 40°C is a rise of 40 K. A Fahrenheit degree is only $\frac{5}{9}$ as big. But a reading on one scale is never the same number as a reading on another, except at the one crossing point each pair of scales has.

Example 1: A weather bulletin in three scales

A summer bulletin reports a peak of 46.0°C in the plains, while a cold-storage warehouse across town is held at $-18.0$°C. Express both on the Fahrenheit and the Kelvin scales, and find the difference between them on all three scales.

Solution:

1. Write the two conversions before any number goes in. $$t_F = \frac{9}{5}\,t_C + 32, \qquad T = t_C + 273.15$$

2. The hot reading. $$t_F = \frac{9}{5}\times 46.0 + 32 = 82.8 + 32 = 114.8°\text{F}$$ $$T = 46.0 + 273.15 = 319.15 \text{ K}$$

3. The cold reading. The minus sign stays with the number all the way through: $$t_F = \frac{9}{5}\times(-18.0) + 32 = -32.4 + 32 = -0.4°\text{F}$$ $$T = -18.0 + 273.15 = 255.15 \text{ K}$$

4. Now the difference — and this is the part worth slowing down for. On the Celsius scale it is $46.0 - (-18.0) = 64.0$ degrees. On the Kelvin scale it is $319.15 - 255.15 = 64.0$ K — the same number, because a kelvin and a Celsius degree are the same size. On the Fahrenheit scale it is $114.8 - (-0.4) = 115.2$ Fahrenheit degrees, which is $\frac{9}{5}\times 64.0$, because a Fahrenheit degree is smaller.

5. The lesson in one line. The $+32$ shifts where zero sits and therefore changes every reading; it cancels out of every difference. Only the factor $\frac{9}{5}$ survives into an interval.

Final Answer: 46.0°C is 114.8°F and 319.15 K; $-18.0$°C is $-0.4$°F and 255.15 K. The gap is 64.0 Celsius degrees, 64.0 K, and 115.2 Fahrenheit degrees.

Takeaway: Converting a reading needs the $+32$; converting an interval does not. Read the question twice and decide which one it is asking for before you write anything.

Example 2: Two cryogenic triple points on three scales

The triple point of hydrogen is 13.81 K and that of oxygen is 54.36 K. Express both on the Celsius and Fahrenheit scales.

Solution:

1. Go through Celsius, because both other conversions are anchored to it. $$t_C = T - 273.15, \qquad t_F = \frac{9}{5}\,t_C + 32$$

2. Hydrogen. $$t_C = 13.81 - 273.15 = -259.34°\text{C}$$ $$t_F = \frac{9}{5}\times(-259.34) + 32 = -466.81 + 32 = -434.81°\text{F}$$

3. Oxygen. $$t_C = 54.36 - 273.15 = -218.79°\text{C}$$ $$t_F = \frac{9}{5}\times(-218.79) + 32 = -393.82 + 32 = -361.82°\text{F}$$

4. Sanity check the size of the answers. Both are far below the coldest natural temperature on Earth, and both Fahrenheit values are below $-400$, which is fine — the Fahrenheit scale only runs out at absolute zero, $-459.67$°F. Any answer below that would be impossible and would tell you the arithmetic had gone wrong.

5. Why these temperatures are quoted in kelvin in the first place. Below about $-200$°C, Celsius numbers are large, negative and awkward, and every gas-law calculation you might want to do with them needs the absolute value anyway. Cryogenics is one field that simply never uses Celsius.

Final Answer: hydrogen $-259.34$°C and $-434.81$°F; oxygen $-218.79$°C and $-361.82$°F.

Takeaway: Route every unfamiliar conversion through Celsius. Two conversions you know beat one you have to reconstruct, and the $-459.67$°F floor is a free check on the answer.

Example 3: Translating between two invented absolute scales

Two absolute temperature scales $P$ and $Q$ are defined by assigning the triple point of water the values 400 P and 250 Q respectively. Find the relation between a temperature $T_P$ on the first scale and $T_Q$ on the second, and check it at the normal boiling point of water, 373.15 K.

Solution:

1. What makes a scale absolute. An absolute scale has its zero at absolute zero, so a reading is simply the kelvin temperature multiplied by a fixed number of scale units per kelvin. That number is fixed by the one assigned point: $$T_P = \left(\frac{400}{273.16}\right)T, \qquad T_Q = \left(\frac{250}{273.16}\right)T$$ where $T$ is in kelvin.

2. Divide one by the other and watch $T$ vanish. $$\frac{T_P}{T_Q} = \frac{400}{250} = 1.6 \qquad \Longrightarrow \qquad T_P = 1.6\,T_Q$$ The relation holds at every temperature, not just at the triple point, precisely because both scales share the same zero.

3. Check it at the boiling point. $$T_P = \frac{400}{273.16}\times 373.15 = 546.4 \text{ P}$$ $$T_Q = \frac{250}{273.16}\times 373.15 = 341.5 \text{ Q}$$ and indeed $\frac{546.4}{341.5} = 1.60$.

4. Why this would fail for Celsius and Fahrenheit. Neither of those has its zero at absolute zero, so their relation carries an offset, $t_F = \frac{9}{5}t_C + 32$, and no simple ratio exists. The clean proportionality here is a direct consequence of the word absolute in the question.

Final Answer: $T_P = 1.6\,T_Q$; at the boiling point of water, 546.4 P and 341.5 Q.

Takeaway: Two absolute scales are always related by a pure multiplying factor, never by a factor plus an offset. Spot the word "absolute" and you have saved yourself half the work.

Example 4: A nickel sensor calibrated at two fixed points

The electrical resistance of a nickel sensor varies with temperature approximately as $R = R_0\left[1 + \kappa\left(T - T_0\right)\right]$. It reads $85.0\ \Omega$ at the triple point of water, 273.16 K, and $152.3\ \Omega$ at the normal boiling point of sulphur, 717.8 K. What temperature does a reading of $118.7\ \Omega$ correspond to?

Solution:

1. Take the triple point as the reference, so $T_0 = 273.16$ K and $R_0 = 85.0\ \Omega$.

2. Find $\kappa$ from the second fixed point. $$152.3 = 85.0\left[1 + \kappa\left(717.8 - 273.16\right)\right]$$ $$\frac{152.3}{85.0} = 1.7918 \qquad \Longrightarrow \qquad \kappa \times 444.64 = 0.7918$$ $$\kappa = 1.781 \times 10^{-3} \text{ K}^{-1}$$

3. Now invert the law for the unknown reading. $$118.7 = 85.0\left[1 + \kappa\left(T - 273.16\right)\right]$$ $$\frac{118.7}{85.0} = 1.3965 \qquad \Longrightarrow \qquad T - 273.16 = \frac{0.3965}{1.781\times 10^{-3}} = 222.66$$ $$T = 273.16 + 222.66 = 495.82 \text{ K}$$

4. A one-line check that needs no $\kappa$ at all. Because the law is linear in $T$, the resistance and the temperature must be at the same fractional position between the two fixed points: $$\frac{118.7 - 85.0}{152.3 - 85.0} = \frac{33.7}{67.3} = 0.5007$$ so $T$ sits almost exactly halfway: $T = 273.16 + 0.5007 \times 444.64 = 495.8$ K. The two routes agree.

5. In Celsius, that is $495.8 - 273.15 = 222.7$°C.

Final Answer: about $495.8$ K, that is $222.7$°C.

Takeaway: A linear thermometric property means proportional interpolation. Once you notice that, you can answer the question by taking the fraction of the way along, and the constant becomes a check rather than a necessity.

Example 5: An unfamiliar scale, and where it agrees with Celsius

A scale $X$ is proposed on which the ice point of water is 10°X and the steam point is 130°X, both at one atmosphere. (a) What does a body at 60°C read on this scale? (b) What Celsius temperature does 46°X correspond to? (c) At what temperature do the Celsius and $X$ scales read the same number? (d) How big is one $X$ degree in Celsius degrees?

Solution:

1. Both scales are linear and share the same two physical fixed points, so the fraction of the way between the fixed points is the same on both: $$\frac{t_X - 10}{130 - 10} = \frac{t_C - 0}{100 - 0}$$

2. Rearrange once and use it for every part. $$t_X = 1.2\,t_C + 10$$

3. (a) At 60°C. $$t_X = 1.2 \times 60 + 10 = 72 + 10 = 82°\text{X}$$

4. (b) At 46°X. $$t_C = \frac{46 - 10}{1.2} = \frac{36}{1.2} = 30°\text{C}$$

5. (c) Where the two agree, set $t_X = t_C = t$: $$t = 1.2t + 10 \qquad \Longrightarrow \qquad -0.2t = 10 \qquad \Longrightarrow \qquad t = -50$$ So $-50$°C is $-50$°X. Every pair of linear scales with different slopes crosses exactly once, and this is where.

6. (d) The size of one degree. The interval from ice to steam is 100 Celsius degrees but 120 $X$ degrees, so $$1°\text{X} = \frac{100}{120} = 0.833 \text{ Celsius degrees}$$ The $X$ degree is the smaller one, which is why 60°C came out as the larger number, 82°X.

Final Answer: (a) 82°X; (b) 30°C; (c) they agree at $-50$ on both scales; (d) one $X$ degree equals $0.833$ Celsius degrees.

Takeaway: Any two-fixed-point scale is fixed by "the same fraction of the way up". Write that sentence as an equation and every part of this question type falls out of one line of algebra.

Example 6: Why helium and nitrogen disagree about zinc

Two constant-volume gas thermometers, one filled with helium and the other with nitrogen, are used to measure the normal freezing point of zinc. The helium thermometer reads $1.100 \times 10^{5}$ Pa at the triple point of water and $2.789 \times 10^{5}$ Pa at the zinc point. The nitrogen thermometer reads $0.2500 \times 10^{5}$ Pa and $0.6345 \times 10^{5}$ Pa at the same two points. (a) What absolute temperature does each give for the zinc point? (b) Why do they disagree, and what would you do about it?

Solution:

1. The defining relation of the constant-volume gas thermometer. With the volume fixed, the pressure of an ideal gas is proportional to the absolute temperature, and the triple point is the single assigned fixed point at 273.16 K: $$T = 273.16 \times \frac{p}{p_{tr}} \ \text{ (kelvin)}$$ This is a ratio of temperatures, so the answer must come out in kelvin — there is no version of this formula in Celsius.

2. (a) The helium thermometer. $$T_A = 273.16 \times \frac{2.789 \times 10^{5}}{1.100 \times 10^{5}} = 273.16 \times 2.5355 = 692.6 \text{ K}$$

3. The nitrogen thermometer. $$T_B = 273.16 \times \frac{0.6345 \times 10^{5}}{0.2500 \times 10^{5}} = 273.16 \times 2.5380 = 693.3 \text{ K}$$

4. The discrepancy. $$T_B - T_A = 693.3 - 692.6 = 0.7 \text{ K}$$ a difference of about one part in a thousand.

5. (b) Why. Neither thermometer is faulty. Helium and nitrogen are real gases, and a real gas obeys $pV = nRT$ only approximately; the deviations differ from gas to gas and grow with the amount of gas present. The ideal-gas law is exact only in the limit of vanishing density.

6. What to do about it. Repeat the measurement with less and less gas in the bulb, so that $p_{tr}$ falls, and plot the deduced temperature against $p_{tr}$. As $p_{tr} \to 0$ the two curves converge on one and the same value, and that common limit is the true thermodynamic temperature. This convergence is exactly why the gas thermometer, and not the mercury thermometer, defines the absolute scale.

Final Answer: (a) 692.6 K from helium and 693.3 K from nitrogen; (b) they differ by 0.7 K because both gases are real; extrapolate both readings to zero pressure and they agree.

Takeaway: The gas thermometer is trusted not because a real gas is ideal but because every real gas becomes ideal as its pressure goes to zero. The extrapolation is the measurement.

Part 2: Expansion of Rods, Plates, Holes and Liquids

Eight problems on $\Delta L = \alpha L\,\Delta T$ and its two relatives. Only $\Delta T$ appears in all three, so Celsius and kelvin are interchangeable throughout this Part — which is exactly why these are the easy marks.

Key Point: $$\Delta L = \alpha L\,\Delta T, \qquad \Delta A = \beta A\,\Delta T, \qquad \Delta V = \gamma V\,\Delta T$$ with $\beta = 2\alpha$ and $\gamma = 3\alpha$ for an isotropic solid. Every length in a heated body — including the diameter of a hole — is multiplied by the same factor $\left(1 + \alpha\,\Delta T\right)$.

Example 7: How much does an overhead line lengthen in summer?

An aluminium conductor is strung in a span 250 m long when its temperature is 15°C. On a summer afternoon the current and the sun together take the conductor to 65°C. Taking $\alpha_{\text{aluminium}} = 2.3 \times 10^{-5}$ K$^{-1}$, how much longer does the conductor become?

Solution:

1. Write the law first. $$\Delta L = \alpha L\,\Delta T$$

2. The temperature change. $\Delta T = 65 - 15 = 50$ degrees, and since only a difference appears, this is 50 K as well — no conversion needed.

3. Substitute. $$\Delta L = 2.3 \times 10^{-5} \times 250 \times 50 = 2.3 \times 10^{-5} \times 12\,500$$ $$\Delta L = 0.2875 \text{ m} = 28.8 \text{ cm}$$

4. Feel the size. Nearly 29 cm of extra conductor in a single span. That is why an overhead line is always strung with a deliberate sag rather than pulled tight: the sag is the slack that lets this expansion happen without the tension climbing until something breaks. Line engineers size the sag for the hottest day of the year, then check the tension on the coldest night.

5. A useful ratio. The fractional change is $\frac{\Delta L}{L} = \alpha\,\Delta T = 1.15 \times 10^{-3}$, about one part in 870. Small — but over 250 m, "small" is still 29 cm.

Final Answer: the conductor lengthens by $0.288$ m, about 29 cm.

Takeaway: A fractional change of $10^{-3}$ sounds negligible until you multiply it by a long object. Always convert $\alpha\,\Delta T$ into an actual length before you decide whether it matters.

Example 8: A glass pane grows in two directions

A window pane measures 1.60 m by 0.90 m at 12°C. Direct sun raises it to 52°C. Taking $\alpha_{\text{glass}} = 9.0 \times 10^{-6}$ K$^{-1}$, find the increase in its area, and show that neglecting the second-order term is justified.

Solution:

1. Areal expansion, with $\beta = 2\alpha$. $$\Delta A = \beta A\,\Delta T = 2\alpha A\,\Delta T$$

2. The starting area and the temperature change. $$A = 1.60 \times 0.90 = 1.44 \text{ m}^2, \qquad \Delta T = 52 - 12 = 40 \text{ K}$$

3. Substitute. $$\Delta A = 2 \times 9.0 \times 10^{-6} \times 1.44 \times 40 = 1.0368 \times 10^{-3} \text{ m}^2$$ which is $10.4$ cm$^2$ — about the area of a large postage stamp.

4. Now earn the approximation. Each side is multiplied by $\left(1 + \alpha\,\Delta T\right)$, so the true new area is $$A^{\,\prime} = A\left(1 + \alpha\,\Delta T\right)^{2} = A\left[1 + 2\alpha\,\Delta T + \left(\alpha\,\Delta T\right)^{2}\right]$$ The term we dropped is $A\left(\alpha\,\Delta T\right)^{2}$. Here $\alpha\,\Delta T = 3.6 \times 10^{-4}$, so $$A\left(\alpha\,\Delta T\right)^{2} = 1.44 \times \left(3.6\times 10^{-4}\right)^{2} = 1.87 \times 10^{-7} \text{ m}^2$$

5. Compare. That is $\frac{1.87 \times 10^{-7}}{1.037 \times 10^{-3}} = 1.8 \times 10^{-4}$, or 0.018% of the answer. The approximation is not a hope; it is a measured 0.018%.

Final Answer: the area grows by $1.04 \times 10^{-3}$ m$^2$, that is $10.4$ cm$^2$. The neglected second-order term is 0.018% of that.

Takeaway: $\beta = 2\alpha$ is not an assumption, it is a binomial expansion whose next term you can actually evaluate. Evaluate it once and you will never worry about it again.

Example 9: How much petrol spills out of a full tanker

A steel road tanker of capacity 12.0 m$^3$ is filled to the brim with petrol at 8°C. During the day the whole tanker and its contents reach 38°C. Taking $\gamma_{\text{petrol}} = 9.5 \times 10^{-4}$ K$^{-1}$ and $\alpha_{\text{steel}} = 1.2 \times 10^{-5}$ K$^{-1}$, how much petrol overflows?

Solution:

1. Both the liquid and its container expand. The spill is the difference, and the difference is governed by the apparent coefficient: $$\text{spill} = V\left(\gamma_{\text{petrol}} - \gamma_{\text{steel}}\right)\Delta T$$

2. Get the tank's volume coefficient from its linear one. $$\gamma_{\text{steel}} = 3\alpha_{\text{steel}} = 3 \times 1.2 \times 10^{-5} = 3.6 \times 10^{-5} \text{ K}^{-1}$$

3. Substitute, with $\Delta T = 30$ K. $$\text{spill} = 12.0 \times \left(9.5 \times 10^{-4} - 3.6 \times 10^{-5}\right)\times 30$$ $$\text{spill} = 12.0 \times 9.14 \times 10^{-4} \times 30 = 0.329 \text{ m}^3 = 329 \text{ litres}$$

4. What the tank's expansion was worth. Had you forgotten it, you would have got $12.0 \times 9.5 \times 10^{-4} \times 30 = 0.342$ m$^3$, that is 342 litres — an error of 13 litres, about 4%. Small here only because steel expands so much less than petrol.

5. Which is exactly why tankers are never filled to the brim. An ullage space, typically a few per cent of the capacity, is left deliberately so that a hot afternoon does not push fuel out of the vent.

Final Answer: about $0.329$ m$^3$, that is 329 litres, spills. Ignoring the tank's own expansion would overstate this by 13 litres.

Takeaway: A liquid in a container always expands by less than you think, because the container grows too. Subtract $3\alpha_{\text{container}}$ from $\gamma_{\text{liquid}}$ before you multiply by anything.

Example 10: A drilled hole, heated two hundred degrees

A hole of diameter 3.60 cm is drilled in an aluminium plate at 15°C. The plate is heated to 215°C. Taking $\alpha_{\text{aluminium}} = 2.3 \times 10^{-5}$ K$^{-1}$, find the change in the diameter of the hole and the change in its area.

Solution:

1. Settle the direction first, because everyone's intuition says the wrong thing. Imagine the disc of metal that was drilled out, put back in place. On heating, the plate and that disc expand identically — they are the same material at the same temperature — so the disc still exactly fills the hole. The hole therefore expands exactly as though it were made of the metal, and it gets bigger.

2. So the hole's diameter obeys the ordinary linear law. $$\Delta d = \alpha\,d\,\Delta T$$

3. Substitute, with $\Delta T = 215 - 15 = 200$ K. $$\Delta d = 2.3 \times 10^{-5} \times 3.60 \times 200 = 1.656 \times 10^{-2} \text{ cm}$$ $$\Delta d = 0.0166 \text{ cm} = 0.166 \text{ mm}$$ so the diameter goes from 3.6000 cm to 3.6166 cm.

4. The area of the hole. Its original area is $$A = \frac{\pi d^{2}}{4} = \frac{\pi \times \left(3.60\right)^{2}}{4} = 10.18 \text{ cm}^2$$ and areal expansion gives $$\Delta A = 2\alpha A\,\Delta T = 2 \times 2.3\times 10^{-5}\times 10.18 \times 200 = 0.0936 \text{ cm}^2$$

5. Check it the long way. The new area is $\frac{\pi\left(3.6166\right)^{2}}{4} = 10.272$ cm$^2$, and $10.272 - 10.179 = 0.093$ cm$^2$. The two agree, as they must.

Final Answer: the hole's diameter increases by $0.166$ mm to $3.6166$ cm; its area increases by $0.0936$ cm$^2$.

Takeaway: A hole is not a thing that can be squeezed — it is a shape, and every shape in a heated body scales up by $\left(1 + \alpha\,\Delta T\right)$. Rivet holes, bearing races and the eye of a needle all get larger when the metal is heated.

Example 11: What a warm steel tape really measured

A steel measuring tape is correctly calibrated at 20°C. On a hot day at 44°C it is used to measure a steel girder and reads 12.500 m. Taking $\alpha_{\text{steel}} = 1.2 \times 10^{-5}$ K$^{-1}$, what is the girder's true length on that day, and what would it be at 20°C?

Solution:

1. Understand what a warm tape does. Every division engraved on the tape has itself grown by the factor $\left(1 + \alpha\,\Delta T\right)$. So each "metre" mark now spans slightly more than a metre, the tape fits fewer times into the girder, and it under-reads. The correction is therefore an addition.

2. The true length on the hot day. $$L_{\text{true}} = L_{\text{read}}\left(1 + \alpha\,\Delta T\right)$$ with $\Delta T = 44 - 20 = 24$ K: $$L_{\text{true}} = 12.500\left(1 + 1.2 \times 10^{-5}\times 24\right) = 12.500\left(1 + 2.88 \times 10^{-4}\right)$$ $$L_{\text{true}} = 12.500 + 0.0036 = 12.5036 \text{ m}$$ The tape was 3.6 mm short over 12.5 m.

3. Now the girder at 20°C. The girder is steel too, so on cooling from 44°C to 20°C it shrinks by exactly the same fractional amount: $$L_{20} = \frac{12.5036}{1 + 1.2\times 10^{-5}\times 24} = 12.5000 \text{ m}$$

4. Read what just happened. When the tape and the object are made of the same material, the tape's reading is automatically the object's length at the tape's calibration temperature — the two errors cancel exactly. That is a genuinely useful piece of workshop physics, and it is also why surveyors' tapes are made of invar, whose $\alpha$ is tiny, when the object being measured is not steel.

5. Say which figure answers which question. On the hot day the girder really is 12.5036 m long. At 20°C it is 12.5000 m long. Both are correct; they are lengths at different temperatures.

Final Answer: $12.5036$ m at 44°C, and $12.5000$ m at 20°C.

Takeaway: A tape that has expanded reads low, so you add the correction. And when the tape and the object share a material, the raw reading is already the length at the calibration temperature.

Example 12: Heating a pulley until it slips onto its shaft

A steel shaft has an outer diameter of 5.000 cm at 25°C. An aluminium pulley whose central hole is 4.988 cm in diameter at the same temperature is to be fitted onto it. To what temperature must the pulley alone be heated so that it just slides on? Take $\alpha_{\text{aluminium}} = 2.3 \times 10^{-5}$ K$^{-1}$.

Solution:

1. The hole must grow to match the shaft, and it grows by the ordinary linear law: $$d\left(1 + \alpha\,\Delta T\right) = d_{\text{shaft}}$$

2. Substitute. $$4.988\left(1 + 2.3 \times 10^{-5}\,\Delta T\right) = 5.000$$ $$1 + 2.3\times 10^{-5}\,\Delta T = \frac{5.000}{4.988} = 1.002406$$

3. Solve for the rise. $$\Delta T = \frac{2.406 \times 10^{-3}}{2.3 \times 10^{-5}} = 104.6 \text{ K}$$

4. The temperature required. $$t = 25 + 104.6 = 129.6°\text{C} \approx 130°\text{C}$$ Comfortably reached in an oil bath or an oven, and well below any temperature that would damage the aluminium.

5. What happens next, and why the fit is so strong. As the pulley cools it tries to shrink back to 4.988 cm but the shaft will not let it, so the pulley is left in permanent tension and grips the shaft with a large radial force. This is a shrink fit — no keyway, no adhesive, just thwarted contraction. The same physics puts an iron tyre on a cartwheel and a steel collar on a rotor.

Final Answer: heat the pulley through $104.6$ K, that is to about $130$°C.

Takeaway: To assemble a shrink fit you can heat the outer part or cool the inner one; the equation is the same, only the sign of $\Delta T$ changes. Solve for $\Delta T$ first and add it to the starting temperature last.

Example 13: What a temperature rise does to a liquid's density

Turpentine has a coefficient of volume expansion $\gamma = 9.4 \times 10^{-4}$ K$^{-1}$ and a density of 870 kg/m$^3$ at 20°C. Find the fractional change in its density and its new density when it is warmed to 70°C.

Solution:

1. Mass is conserved; volume is not. With $\rho = \frac{m}{V}$ and $V$ growing to $V\left(1 + \gamma\,\Delta T\right)$, $$\rho^{\,\prime} = \frac{m}{V\left(1 + \gamma\,\Delta T\right)} = \frac{\rho}{1 + \gamma\,\Delta T}$$

2. The fractional change, to first order. Expanding the denominator, $$\frac{\Delta \rho}{\rho} \approx -\gamma\,\Delta T = -9.4 \times 10^{-4}\times 50 = -0.047$$ a decrease of 4.7%. The minus sign is the physics: heating a liquid always makes it less dense — with water between 0°C and 4°C as the famous exception.

3. The new density, done exactly. $$\rho^{\,\prime} = \frac{870}{1 + 0.047} = \frac{870}{1.047} = 830.9 \text{ kg/m}^3$$

4. The new density, done to first order. $$\rho^{\,\prime} \approx 870\left(1 - 0.047\right) = 829.1 \text{ kg/m}^3$$ The two differ by 1.8 kg/m$^3$, about 0.2% — acceptable for most purposes, but note that the exact route is no harder, so use it.

5. Why this matters commercially. Fuels and edible oils are bought by volume but priced on mass, so every delivery is corrected to a standard temperature, usually 15°C. A 4.7% density change is a 4.7% billing error.

Final Answer: the density falls by $4.7\%$, from 870 to $830.9$ kg/m$^3$.

Takeaway: $\frac{\Delta\rho}{\rho} = -\gamma\,\Delta T$ — same magnitude as the volume change, opposite sign. When the change is more than a per cent or two, divide rather than subtract.

Example 14: Overflow from a steel vessel, and the coefficient it hides

A steel vessel of capacity 500 cm$^3$ is filled to the brim with glycerine at 20°C. When the whole thing is heated to 80°C, $13.5$ cm$^3$ of glycerine overflows. Taking $\alpha_{\text{steel}} = 1.2 \times 10^{-5}$ K$^{-1}$, find the apparent and the real coefficient of volume expansion of glycerine.

Solution:

1. What the overflow actually measures. The spilt volume is the amount by which the liquid out-expands its container, so it gives the apparent coefficient directly: $$\text{spill} = V\,\gamma_{app}\,\Delta T \qquad \Longrightarrow \qquad \gamma_{app} = \frac{\text{spill}}{V\,\Delta T}$$

2. Substitute, with $\Delta T = 60$ K. $$\gamma_{app} = \frac{13.5}{500 \times 60} = \frac{13.5}{30\,000} = 4.5 \times 10^{-4} \text{ K}^{-1}$$

3. Add back what the vessel did. $$\gamma_{real} = \gamma_{app} + \gamma_{\text{steel}} = \gamma_{app} + 3\alpha_{\text{steel}}$$ $$\gamma_{real} = 4.5 \times 10^{-4} + 3.6 \times 10^{-5} = 4.86 \times 10^{-4} \text{ K}^{-1}$$

4. Check the answer against reality. Glycerine's accepted coefficient of volume expansion is about $4.9 \times 10^{-4}$ K$^{-1}$, so the experiment has landed within about 1% — which is roughly the accuracy such a measurement deserves.

5. The order of operations matters. You measure the apparent value and add the vessel's contribution to get the real one. Doing it the other way round — subtracting — is a standard slip, and gives $4.14 \times 10^{-4}$, which is 15% low.

Final Answer: $\gamma_{app} = 4.5 \times 10^{-4}$ K$^{-1}$ and $\gamma_{real} = 4.86 \times 10^{-4}$ K$^{-1}$.

Takeaway: The overflow only ever measures the difference $\gamma_{real} - \gamma_{vessel}$. Whatever a problem gives you about the container, put it back in before you claim to have found a property of the liquid.

Part 3: Expansion in Service — Joints, Fits, Strips and Stress

Six problems in which expansion is either allowed for or forbidden. When it is forbidden, the force that appears is enormous, and that force is the whole point.

Key Point: A rod that is heated but not allowed to expand develops a compressive stress $$\text{stress} = Y\alpha\,\Delta T$$ independent of its length and its area. The force on the supports, $F = YA\alpha\,\Delta T$, does depend on the area. Cool a clamped rod instead and the same formula gives a tension.

Example 15: The tension a cooled wire pulls with

A copper wire of diameter 1.5 mm is stretched with negligible tension between two rigid supports 2.4 m apart at 30°C. It is then cooled to $-20$°C. Taking $\alpha_{\text{copper}} = 1.7 \times 10^{-5}$ K$^{-1}$ and Young's modulus $Y_{\text{copper}} = 1.1 \times 10^{11}$ Pa, find the tension that develops.

Solution:

1. The wire wants to contract but cannot. Free contraction would give a strain $$\frac{\Delta L}{L} = \alpha\,\Delta T$$ The supports hold the ends apart, so this strain is exactly cancelled by an elastic tensile strain of the same size.

2. Turn that strain into a stress with Young's modulus. $$\text{stress} = Y \times \text{strain} = Y\alpha\,\Delta T$$ $$\text{stress} = 1.1 \times 10^{11} \times 1.7\times 10^{-5}\times 50 = 9.35 \times 10^{7} \text{ Pa}$$ using $\Delta T = 30 - (-20) = 50$ K.

3. Find the cross-sectional area. The radius is $0.75$ mm $= 7.5 \times 10^{-4}$ m: $$A = \pi r^{2} = \pi\left(7.5 \times 10^{-4}\right)^{2} = 1.767 \times 10^{-6} \text{ m}^2$$

4. Multiply. $$F = \text{stress} \times A = 9.35 \times 10^{7} \times 1.767 \times 10^{-6} = 165 \text{ N}$$

5. Notice what never entered. The length 2.4 m appears nowhere in the answer. A wire twice as long would contract twice as much in absolute terms, but it also has twice as much length over which to stretch, so the strain — and therefore the stress and the tension — is unchanged. Length is a distractor in every problem of this shape.

Final Answer: a tension of about $165$ N, corresponding to a tensile stress of $9.35 \times 10^{7}$ Pa.

Takeaway: Thermal stress depends on $Y$, $\alpha$ and $\Delta T$ and on nothing else; the force additionally needs the area. If a problem hands you a length, check whether it is doing any work before you use it.

Example 16: Two rods end to end, free to grow

An aluminium rod 40.0 cm long is joined end to end to a copper rod 60.0 cm long, both of diameter 4.0 mm. The pair is heated from 20°C to 180°C with both outer ends free to move. Taking $\alpha_{\text{aluminium}} = 2.3 \times 10^{-5}$ K$^{-1}$ and $\alpha_{\text{copper}} = 1.7 \times 10^{-5}$ K$^{-1}$, find the change in the total length. Is there a thermal stress at the junction?

Solution:

1. Each rod expands according to its own $\alpha$ and its own length. They are in series, so their extensions simply add: $$\Delta L_{\text{total}} = \alpha_{Al}L_{Al}\,\Delta T + \alpha_{Cu}L_{Cu}\,\Delta T$$

2. The temperature change is $\Delta T = 180 - 20 = 160$ K.

3. The aluminium. $$\Delta L_{Al} = 2.3 \times 10^{-5}\times 0.400 \times 160 = 1.472 \times 10^{-3} \text{ m} = 1.472 \text{ mm}$$

4. The copper. $$\Delta L_{Cu} = 1.7 \times 10^{-5}\times 0.600 \times 160 = 1.632 \times 10^{-3} \text{ m} = 1.632 \text{ mm}$$

5. Add. $$\Delta L_{\text{total}} = 1.472 + 1.632 = 3.104 \text{ mm}$$ so the combined rod goes from $1.000$ m to $1.003104$ m. Notice that the copper contributes more even though its $\alpha$ is smaller, because it is half as long again.

6. Is there a stress at the junction? No. Stress requires that something be prevented from reaching its natural length, and here nothing is: both ends are free, so each rod expands to exactly the length it wants and the junction simply moves along. The joint has to carry no force at all. Clamp one end and the answer changes completely — but the question says free.

7. The diameter, 4.0 mm, is not needed for the length question. It would be needed only if a force were asked for, and there is no force.

Final Answer: the combined rod lengthens by $3.10$ mm; there is no thermal stress at the junction because both ends are free.

Takeaway: Free ends mean zero thermal stress, however different the two materials are. Stress appears only when a rigid constraint refuses the expansion — look for the word "clamped", "rigid" or "fixed" before you reach for $Y\alpha\,\Delta T$.

Example 17: When does the brass rod overtake the steel one?

At 0°C a brass rod and a steel rod are each exactly 1.200 m long. Both are heated through the same temperature rise. At what temperature is the brass rod 1.0 mm longer than the steel one? Take $\alpha_{\text{brass}} = 1.8 \times 10^{-5}$ K$^{-1}$ and $\alpha_{\text{steel}} = 1.2 \times 10^{-5}$ K$^{-1}$.

Solution:

1. Write both lengths at temperature $t$, measured from the common starting point of 0°C: $$L_{brass} = L\left(1 + \alpha_{b}\,t\right), \qquad L_{steel} = L\left(1 + \alpha_{s}\,t\right)$$

2. Subtract, and watch the common length factor out. $$L_{brass} - L_{steel} = L\left(\alpha_{b} - \alpha_{s}\right)t$$ This is the key structural fact: the difference in length grows in proportion to the difference in the coefficients.

3. Substitute the required gap of 1.0 mm. $$1.0 \times 10^{-3} = 1.200 \times \left(1.8 \times 10^{-5} - 1.2 \times 10^{-5}\right)\times t$$ $$1.0 \times 10^{-3} = 1.200 \times 0.6 \times 10^{-5}\times t = 7.2 \times 10^{-6}\,t$$

4. Solve. $$t = \frac{1.0 \times 10^{-3}}{7.2 \times 10^{-6}} = 138.9°\text{C}$$

5. Check the scale of it. Nearly 140 degrees to open a gap of one millimetre between two rods more than a metre long. Differential expansion is a small effect — which is exactly why it needs a lever, or a long strip, or a fine screw to be made visible. That is the next problem.

Final Answer: at about $139$°C, that is a rise of 139 K above the starting point.

Takeaway: Differential expansion depends on $\left(\alpha_1 - \alpha_2\right)$, not on either $\alpha$ separately. Two metals with a large gap between their coefficients make a sensitive device; two with similar coefficients make none at all.

Example 18: How hot before a fire-alarm strip closes its contact?

A bimetallic strip is made of a brass strip and an iron strip, each 8.0 cm long, bonded face to face and straight at 25°C. The alarm is set to trip when the difference between the free lengths of the two metals would reach 30 micrometres. Taking $\alpha_{\text{brass}} = 1.8 \times 10^{-5}$ K$^{-1}$ and $\alpha_{\text{iron}} = 1.2 \times 10^{-5}$ K$^{-1}$, at what temperature does the alarm trip, and which way does the strip bend?

Solution:

1. The mechanism, in one sentence. The two metals are bonded, so they must stay the same length; but on heating brass wants to be longer than iron. The only way to satisfy both is for the strip to curve, with the brass on the outside of the arc where there is more room.

2. So the direction is settled before any arithmetic. The strip bends towards the iron, the metal with the smaller $\alpha$. Cool it below 25°C instead and it bends the other way, towards the brass.

3. The quantity that drives the bend is the difference in free lengths: $$\delta = \left(\alpha_{brass} - \alpha_{iron}\right)L\,\Delta T$$

4. Set it equal to the trip value $30 \times 10^{-6}$ m. $$30 \times 10^{-6} = \left(1.8\times 10^{-5} - 1.2\times 10^{-5}\right)\times 0.080 \times \Delta T$$ $$30 \times 10^{-6} = 4.8 \times 10^{-7}\,\Delta T$$

5. Solve for the rise and add it on. $$\Delta T = \frac{30 \times 10^{-6}}{4.8 \times 10^{-7}} = 62.5 \text{ K}$$ $$t_{\text{trip}} = 25 + 62.5 = 87.5°\text{C}$$

6. Why this is a good alarm. The trip point is set purely by the geometry and the two coefficients — no electronics, no power supply, nothing to fail. The same strip, run the other way, is the thermostat in an iron, a geyser and an oven, and it was the flasher in every old indicator lamp.

Final Answer: the alarm trips at about $87.5$°C, and the strip bends towards the iron.

Takeaway: A bimetallic strip always curves towards the metal with the smaller $\alpha$ when heated, and away from it when cooled. Decide the direction from that sentence, then do the arithmetic.

Example 19: The force a clamped girder pushes with

A steel girder of cross-sectional area 45 cm$^2$ is fitted between two rigid abutments at 10°C. In summer its temperature rises to 42°C. Taking $\alpha_{\text{steel}} = 1.2 \times 10^{-5}$ K$^{-1}$ and $Y_{\text{steel}} = 2.0 \times 10^{11}$ Pa, find the thermal stress and the force it exerts on each abutment.

Solution:

1. The girder is prevented from expanding, so it is compressed by exactly the amount it wanted to grow. $$\text{strain} = \alpha\,\Delta T = 1.2\times 10^{-5}\times 32 = 3.84 \times 10^{-4}$$ with $\Delta T = 42 - 10 = 32$ K.

2. Convert to stress. $$\text{stress} = Y\alpha\,\Delta T = 2.0 \times 10^{11}\times 3.84 \times 10^{-4} = 7.68 \times 10^{7} \text{ Pa}$$

3. Convert the area to SI before multiplying. $$A = 45 \text{ cm}^2 = 45 \times 10^{-4} \text{ m}^2 = 4.5 \times 10^{-3} \text{ m}^2$$

4. The force. $$F = \text{stress} \times A = 7.68 \times 10^{7} \times 4.5 \times 10^{-3} = 3.46 \times 10^{5} \text{ N}$$

5. Put that in human terms. $3.46 \times 10^{5}$ N is the weight of about 35 tonnes — roughly two loaded trucks pressing on each abutment, produced by nothing more than a 32-degree change in the weather. The stress itself, 77 MPa, is a serious fraction of structural steel's yield strength of around 250 MPa, so it cannot be shrugged off.

6. Which is why bridges have expansion joints. Let the steel move a few millimetres and the stress is zero; refuse it those millimetres and you get 35 tonnes.

Final Answer: a compressive stress of $7.68 \times 10^{7}$ Pa and a force of $3.46 \times 10^{5}$ N on each abutment.

Takeaway: Thermal stress is savage because $Y$ is huge. A strain of only $4 \times 10^{-4}$ becomes 77 MPa once it is multiplied by a Young's modulus of $2 \times 10^{11}$ Pa.

Example 20: Sizing the loop in a steam pipe

A straight run of steel steam pipe 42 m long is installed at 22°C. In service it carries steam at 180°C. Taking $\alpha_{\text{steel}} = 1.2 \times 10^{-5}$ K$^{-1}$, how much movement must the expansion loop at the end of the run absorb?

Solution:

1. The loop has to swallow exactly the free expansion of the straight run. $$\Delta L = \alpha L\,\Delta T$$

2. The temperature change. $\Delta T = 180 - 22 = 158$ K. Note that this is the temperature of the pipe metal, which reaches the steam temperature, not the temperature of the room.

3. Substitute. $$\Delta L = 1.2 \times 10^{-5}\times 42 \times 158 = 1.2 \times 10^{-5}\times 6636$$ $$\Delta L = 0.0796 \text{ m} = 7.96 \text{ cm}$$

4. Design the loop for more than that, not exactly that. A designer would allow perhaps 12 cm — the pipe may be commissioned on a cold night rather than at 22°C, and a loop that runs out of travel transmits the full $Y\alpha\,\Delta T$ stress straight into the pipe flanges and the anchor points.

5. Compare with what would happen without a loop. Clamp the run rigidly and the stress would be $$Y\alpha\,\Delta T = 2.0 \times 10^{11}\times 1.2\times 10^{-5}\times 158 = 3.8 \times 10^{8} \text{ Pa}$$ which is well past the yield strength of ordinary structural steel. The pipe would buckle. Eight centimetres of designed-in slack is what stands between the plant and that.

Final Answer: the loop must absorb about $8.0$ cm of movement.

Takeaway: Every long hot run of metal needs somewhere to put its expansion — a loop, a joint, a gap or a sliding support. Compute $\alpha L\,\Delta T$, then design in more than that.

Part 4: Specific Heat, Molar Heat, and Work Turned into Heat

Six problems on $Q = ms\,\Delta T$ and its cousins. Again only $\Delta T$ appears, so Celsius and kelvin are interchangeable — except in the one problem that starts in Fahrenheit, where you must convert before you subtract.

Key Point: $$Q = m\,s\,\Delta T \quad \text{(per kilogram)}, \qquad Q = \mu\,C\,\Delta T \quad \text{(per mole)}, \qquad S = m\,s \quad \text{(whole body)}$$ $s$ is in J/(kg K), $C$ in J/(mol K), $S$ in J/K. When several bodies are heated together, add their $ms$ products — that total is the heat capacity of the whole arrangement.

Example 21: Heating the pot as well as the water

An aluminium pan of mass 0.60 kg holds 1.8 kg of water at 24°C and is put on a 2.0 kW hob. Taking $s_{\text{aluminium}} = 900$ J/(kg K) and $s_{\text{water}} = 4186$ J/(kg K), how long does it take to reach 95°C if no heat is lost? What fraction of the energy goes into the pan?

Solution:

1. Both the pan and the water are heated through the same $\Delta T$, so add their heat capacities: $$Q = \left(m_{pan}s_{pan} + m_{water}s_{water}\right)\Delta T$$

2. Compute each heat capacity separately — it is worth seeing the two numbers side by side. $$m_{pan}s_{pan} = 0.60 \times 900 = 540 \text{ J/K}$$ $$m_{water}s_{water} = 1.8 \times 4186 = 7534.8 \text{ J/K}$$

3. The heat required, with $\Delta T = 95 - 24 = 71$ K: $$Q = \left(540 + 7534.8\right)\times 71 = 8074.8 \times 71 = 5.733 \times 10^{5} \text{ J}$$

4. The time. $$t = \frac{Q}{P} = \frac{5.733 \times 10^{5}}{2000} = 287 \text{ s} = 4.8 \text{ minutes}$$

5. The pan's share. $$\frac{540}{8074.8} = 0.067 = 6.7\%$$ Only about a fifteenth of the gas bill goes into the metal. Water's enormous specific heat capacity dominates so completely that ignoring the pan would have cost you only 6.7% — but in a calorimetry experiment, where you are chasing 1% accuracy, 6.7% is a disaster. That is precisely why a calorimeter's own heat capacity is always accounted for.

Final Answer: about $287$ s, that is $4.8$ minutes; the pan takes $6.7\%$ of the energy.

Takeaway: When several bodies are heated together, add their $ms$ products, not their masses. And always check what fraction the "small" body takes — sometimes it is negligible and sometimes it decides the answer.

Example 22: Reading a table of molar specific heats

The table below gives measured molar specific heats at constant volume, at room temperature, for several common gases. A monatomic gas such as helium or argon measures about 12.2 J/(mol K). Explain the difference, and comment on chlorine's noticeably larger value. Take $R = 8.31$ J/(mol K).

Gas	$C_v$ measured, J/(mol K)
Hydrogen	20.4
Nitrogen	20.8
Oxygen	21.0
Nitric oxide	20.9
Carbon monoxide	21.0
Chlorine	25.8
A monatomic gas, for comparison	12.2

Measured values for common gases near room temperature, rounded to three figures. In this problem $R$ is the gas constant, not a thermal resistance.

Solution:

1. Start with the monatomic gas, which is the simplest case. A single atom can only move — three independent directions of translation, and nothing else. Each such degree of freedom carries $\frac{1}{2}R$ per mole, so $$C_v = \frac{3}{2}R = \frac{3}{2}\times 8.31 = 12.5 \text{ J/(mol K)}$$ which matches the measured 12.2 nicely.

2. Now a diatomic molecule. It is a dumbbell, so besides moving through space it can also rotate — about two axes perpendicular to the bond. Rotation about the bond itself carries no appreciable energy. That gives two extra degrees of freedom, so $$C_v = \frac{5}{2}R = \frac{5}{2}\times 8.31 = 20.8 \text{ J/(mol K)}$$

3. Compare with the table. Hydrogen 20.4, nitrogen 20.8, oxygen 21.0, nitric oxide 20.9, carbon monoxide 21.0. Every one of them is a diatomic molecule, and every one of them sits right on $\frac{5}{2}R$. The agreement is the point: the extra heat a diatomic gas needs is the energy going into rotation.

4. Now chlorine, at 25.8. That is well above $\frac{5}{2}R$ and heading towards $\frac{7}{2}R = 29.1$ J/(mol K), which is what you get by adding a vibrational mode — the two atoms oscillating along the bond, contributing both kinetic and potential energy, so $R$ rather than $\frac{1}{2}R$.

5. Why chlorine and not nitrogen? Chlorine's atoms are heavy and its bond is comparatively weak, so its vibration is slow and low in energy — low enough to be appreciably excited at room temperature. Nitrogen's bond is a strong triple bond between light atoms, so its vibration needs far more energy and stays frozen out until the gas is very hot. The measured $C_v$ is therefore a direct readout of which internal motions a molecule has switched on.

Final Answer: the diatomic gases sit at $\frac{5}{2}R \approx 20.8$ J/(mol K) because they rotate as well as translate, against $\frac{3}{2}R \approx 12.5$ for a monatomic gas; chlorine exceeds this because its low-frequency vibration is already partly excited at room temperature.

Takeaway: Count the ways a molecule can hold energy and you have predicted its molar specific heat. A number above $\frac{5}{2}R$ for a diatomic gas is a signature of vibration.

Example 23: A drill, a copper block and three minutes

A 6.0 kW drilling machine is used on a copper block of mass 20 kg for 3.0 minutes. Sixty per cent of the machine's power ends up heating the block, the rest going into the machine itself and the surroundings. Taking $s_{\text{copper}} = 386.4$ J/(kg K), find the rise in the block's temperature.

Solution:

1. Mechanical work done against friction reappears as heat. The useful power is $$P_{\text{useful}} = 0.60 \times 6.0 \times 10^{3} = 3.6 \times 10^{3} \text{ W}$$

2. The energy delivered in 3.0 minutes. Convert the time to seconds first — $3.0$ min $= 180$ s: $$Q = P_{\text{useful}}\,t = 3.6 \times 10^{3}\times 180 = 6.48 \times 10^{5} \text{ J}$$

3. Turn that into a temperature rise. $$Q = m s\,\Delta T \qquad \Longrightarrow \qquad \Delta T = \frac{Q}{m s}$$ $$\Delta T = \frac{6.48 \times 10^{5}}{20 \times 386.4} = \frac{6.48 \times 10^{5}}{7728} = 83.9 \text{ K}$$

4. Where the block ends up. Starting from a workshop temperature of about 30°C, the block would reach roughly 114°C — hot enough to burn a hand and hot enough to matter for the tool's cutting edge. This is why deep drilling uses a cutting fluid: it carries the heat away by convection instead of letting it pile up in the workpiece.

5. Watch the two traps in this problem. First, the 40% that is lost never enters the block, so it must be removed before you multiply by the time. Second, minutes are not seconds; forgetting that gives an answer 60 times too small.

Final Answer: the block warms by about $84$ K.

Takeaway: Work against friction is a heat source like any other; the only new step is deciding what fraction of the power actually lands in the body you care about. Apply the efficiency before the time, not after.

Example 24: How much warmer is the water at the foot of a waterfall?

Water falls 105 m over a waterfall. Assuming all the kinetic energy gained in the fall turns into internal energy of the water itself, by how much does the water warm up? Take $g = 9.8$ m/s$^2$ and $s_{\text{water}} = 4186$ J/(kg K).

Solution:

1. Work out the energy per kilogram, because the mass will cancel and it is cleaner to see that in advance. A mass $m$ falling through height $h$ loses potential energy $$\Delta U = m g h$$

2. Set that equal to the heat that appears in the same mass of water. $$m g h = m s\,\Delta T$$

3. Cancel the mass and solve. $$\Delta T = \frac{g h}{s}$$

4. Substitute. $$\Delta T = \frac{9.8 \times 105}{4186} = \frac{1029}{4186} = 0.246 \text{ K}$$

5. Read the answer honestly. A quarter of a degree, from a 105 m fall — and even that is an over-estimate, because in reality a good deal of the energy goes into churning the air, into sound and into spray that evaporates. Joule famously looked for exactly this effect on his honeymoon and could not measure it reliably. The reason it is so small is water's specific heat capacity: 4186 J will raise one kilogram by one kelvin, and one kilogram falling 105 m only releases about 1000 J.

6. The scaling is worth storing. You would need a fall of roughly 427 m to warm the water by a full kelvin.

Final Answer: about $0.25$ K.

Takeaway: Gravity is a feeble heater because $s$ for water is enormous. Whenever mechanical energy is converted to heat in water, expect an answer in fractions of a degree unless the speeds are very large.

Example 25: The sweat that brings a fever down

A child of mass 28 kg has a temperature of $102.2$°F. A medicine is given that works by increasing the rate of evaporation of sweat, and the temperature falls to $99.5$°F in 25 minutes. Assuming evaporation is the only route by which heat leaves, and taking the specific heat capacity of the human body as that of water, 4186 J/(kg K), with the latent heat of evaporation of water at body temperature as $2.42 \times 10^{6}$ J/kg, find the average extra rate of evaporation caused by the drug.

Solution:

1. Convert both temperatures to Celsius before subtracting. This is the one place in this Part where the conversion matters, because a Fahrenheit degree is not the same size as a kelvin. $$t_C = \frac{5}{9}\left(t_F - 32\right)$$ $$102.2°\text{F} \to \frac{5}{9}\left(70.2\right) = 39.0°\text{C}, \qquad 99.5°\text{F} \to \frac{5}{9}\left(67.5\right) = 37.5°\text{C}$$ So the fall is $\Delta T = 1.5$ K. (Doing it wrong — subtracting the Fahrenheit readings to get $2.7$ and calling that kelvin — would inflate the answer by 80%.)

2. The heat that had to leave the body. $$Q = m s\,\Delta T = 28 \times 4186 \times 1.5 = 1.758 \times 10^{5} \text{ J}$$

3. All of it is carried away as latent heat of evaporation. $$Q = m_{\text{sweat}}L_v \qquad \Longrightarrow \qquad m_{\text{sweat}} = \frac{1.758 \times 10^{5}}{2.42 \times 10^{6}}$$ $$m_{\text{sweat}} = 0.0727 \text{ kg} = 72.7 \text{ g}$$

4. The rate. $$\frac{72.7 \text{ g}}{25 \text{ min}} = 2.91 \text{ g per minute}$$ or $4.8 \times 10^{-5}$ kg/s.

5. Sanity check the physiology. About 73 g of sweat — a small glassful — over 25 minutes. That is entirely plausible, and it explains why a feverish patient must be kept hydrated: the cooling mechanism is literally spending the body's water.

Final Answer: about $72.7$ g of sweat, an average rate of $2.9$ g per minute.

Takeaway: Evaporation is a spectacularly efficient coolant because $L_v$ is huge. Evaporating one gram of water removes as much heat as cooling 578 grams of water by one kelvin.

Example 26: How long does the geyser take?

A 2.0 kW immersion geyser heats 20 litres of water from 18°C to 55°C. If 12% of the electrical energy is lost to the surroundings and the tank, how long does it take? Take the density of water as 1000 kg/m$^3$ and $s_{\text{water}} = 4186$ J/(kg K).

Solution:

1. Convert the volume to a mass, since $Q = ms\,\Delta T$ needs a mass. Twenty litres is $20 \times 10^{-3}$ m$^3$, so $$m = 1000 \times 20 \times 10^{-3} = 20 \text{ kg}$$ For water, litres and kilograms are numerically the same, which is convenient but worth stating rather than assuming.

2. The heat the water actually needs. $$Q = m s\,\Delta T = 20 \times 4186 \times \left(55 - 18\right) = 20 \times 4186 \times 37$$ $$Q = 3.098 \times 10^{6} \text{ J}$$

3. The power that reaches the water. Only 88% of the input arrives: $$P_{\text{useful}} = 0.88 \times 2000 = 1760 \text{ W}$$

4. The time. $$t = \frac{Q}{P_{\text{useful}}} = \frac{3.098 \times 10^{6}}{1760} = 1.76 \times 10^{3} \text{ s}$$ which is $29$ minutes.

5. What the 12% cost. With no losses at all the time would have been $\frac{3.098\times 10^{6}}{2000} = 1549$ s, that is 26 minutes. The losses added about three and a half minutes and, over a year, a noticeable amount to the bill — which is the whole argument for lagging a hot-water tank.

6. A useful rule of thumb worth remembering. Raising one litre of water by one kelvin needs about 4.2 kJ. So a 2 kW heater raises one litre by about half a kelvin per second, and 20 litres by about 0.025 K/s. Multiply by 37 K and you get about 1500 s before losses — the answer, near enough, without a calculator.

Final Answer: about $1.76 \times 10^{3}$ s, that is 29 minutes.

Takeaway: Efficiency reduces the power, it does not reduce the heat required. Compute $Q$ from the water alone, then divide by the power that actually gets there.

Part 5: Calorimetry, With and Without a Phase Change

Seven problems built on one equation — heat lost equals heat gained — and one habit that separates a correct answer from a wrong one: never assume all the ice melts.

Key Point — the branch test, and do it every single time: Before writing a final temperature, compute two numbers separately:

• the heat available, that is how much the warm bodies can give up before they reach the change-of-state temperature;
• the heat required to complete the change of state.

If available $\geq$ required, the change of state finishes and the final temperature moves past it. If available $<$ required, the change of state stops part-way and the final temperature pins itself at the change-of-state temperature, with a mixture left behind. Say which branch you took, in words, in your answer.

Example 27: Tea, tumbler and final temperature

A steel tumbler of mass 0.180 kg is at 22°C. Into it is poured 0.220 kg of tea at 88°C. Taking $s_{\text{steel}} = 450$ J/(kg K) and the tea's specific heat capacity as that of water, 4186 J/(kg K), and assuming no heat escapes, find the final temperature.

Solution:

1. Branch test first. The only phase change available would be boiling or freezing, and nothing here is near 100°C or 0°C. So no phase change is possible and we can go straight to a temperature balance.

2. Write the balance. The tea loses heat, the tumbler gains it: $$m_{tea}s_{water}\left(88 - T\right) = m_{cup}s_{steel}\left(T - 22\right)$$

3. Compute the two heat capacities before substituting — it keeps the algebra clean. $$m_{tea}s_{water} = 0.220 \times 4186 = 920.9 \text{ J/K}$$ $$m_{cup}s_{steel} = 0.180 \times 450 = 81.0 \text{ J/K}$$

4. Substitute and solve. $$920.9\left(88 - T\right) = 81.0\left(T - 22\right)$$ $$81\,041 - 920.9\,T = 81.0\,T - 1782$$ $$82\,823 = 1001.9\,T \qquad \Longrightarrow \qquad T = 82.7°\text{C}$$

5. Check that the answer sits between the two starting temperatures. It does — 82.7 lies between 22 and 88 — and it sits much closer to the tea's temperature, which is right, because the tea's heat capacity is more than eleven times the tumbler's. Any answer outside the range 22 to 88 would be impossible, and that check costs two seconds.

6. The physical reading. The tea drops only 5.3 K. A steel tumbler steals surprisingly little heat, because steel's specific heat capacity is about one ninth of water's and the masses are comparable.

Final Answer: $82.7$°C — the tea cools by $5.3$ K.

Takeaway: Compute $ms$ for every body first, then write one equation. The body with the larger $ms$ always drags the final temperature towards itself, and that gives you a free check on the answer.

Example 28: Pinning down an unknown metal's specific heat

A 0.15 kg block of an unknown metal at 180°C is dropped into a calorimeter of water equivalent 0.020 kg containing 200 cm$^3$ of water at 25°C. The final temperature is $36.5$°C. Taking $s_{\text{water}} = 4186$ J/(kg K), find the specific heat capacity of the metal. If some heat did leak away, is the answer too big or too small?

Solution:

1. What "water equivalent" buys you. The water equivalent $W$ of the calorimeter is the mass of water that would absorb the same heat for the same temperature rise. So the calorimeter can be folded into the arithmetic simply as 0.020 kg of extra water — no separate term needed.

2. The mass of water. 200 cm$^3$ of water is $200$ g $= 0.200$ kg.

3. The heat gained by the water plus calorimeter. $$Q_{\text{gained}} = \left(m_{water} + W\right)s_{water}\left(36.5 - 25\right)$$ $$Q_{\text{gained}} = \left(0.200 + 0.020\right)\times 4186 \times 11.5 = 0.220 \times 4186 \times 11.5 = 1.059 \times 10^{4} \text{ J}$$

4. Set that equal to the heat lost by the metal. $$m\,s\left(180 - 36.5\right) = 1.059 \times 10^{4}$$ $$0.15 \times s \times 143.5 = 1.059 \times 10^{4} \qquad \Longrightarrow \qquad 21.53\,s = 1.059\times 10^{4}$$ $$s = 492 \text{ J/(kg K)}$$

5. Does the answer make sense? About 492 J/(kg K) is in the region of iron and steel, well above copper's 386 and well below aluminium's 900. A plausible identification.

6. Now the error question, which is worth reasoning out rather than guessing. If heat leaks to the surroundings, the metal really gave out more heat than the water and calorimeter received. Our equation credited the metal with only what the water gained, so it under-counted the heat the metal supplied, and therefore under-counted $s$. The measured value is smaller than the true one.

7. How the experiment fights back. Start the water a few degrees below room temperature and finish it a few degrees above, so that the leak runs inwards for the first half and outwards for the second, and the two roughly cancel. A vacuum flask does the same job by brute force.

Final Answer: $s \approx 492$ J/(kg K); with heat losses, this is an under-estimate of the true value.

Takeaway: Water equivalent turns the calorimeter into extra water, and that is the whole reason the quantity exists. And a heat leak always makes the hot body look as though it had less heat to give, so a measured specific heat comes out low.

Example 29: How much ice can a hot iron block melt?

A block of iron of mass 1.8 kg is heated to 400°C and then placed on a large block of ice at 0°C. Taking $s_{\text{iron}} = 450$ J/(kg K) and $L_f = 3.33 \times 10^{5}$ J/kg, what is the greatest mass of ice that can melt?

Solution:

1. Branch test. The ice block is described as large, which is doing real work in this problem: it means there is more ice available than the block can possibly melt. So the ice never all melts, the final temperature pins at 0°C, and this is the partial-melt branch.

2. The most heat the iron can give up is what it releases in cooling all the way to 0°C — it cannot go below the temperature of the ice it is sitting on. $$Q = m s\,\Delta T = 1.8 \times 450 \times \left(400 - 0\right)$$ $$Q = 1.8 \times 450 \times 400 = 3.24 \times 10^{5} \text{ J}$$

3. Turn that into melted ice. $$Q = m_{ice}L_f \qquad \Longrightarrow \qquad m_{ice} = \frac{3.24 \times 10^{5}}{3.33 \times 10^{5}}$$ $$m_{ice} = 0.973 \text{ kg}$$

4. Read the answer. Not quite one kilogram of ice — from a block of iron heated almost to visible red heat. Nearly a kilogram of ice absorbs as much energy in melting as 1.8 kg of iron gives up in falling 400 degrees. That single comparison is the best argument for how large $L_f$ really is.

5. Why the word "maximum" is in the question. In practice some of the heat warms the air and the puddle of meltwater, so less ice melts. The calculation gives the ceiling.

Final Answer: at most $0.973$ kg, about $973$ g of ice. Branch taken: partial melt, final temperature 0°C.

Takeaway: "A large block of ice" is a statement about the branch, not about the geometry. It tells you the final temperature is 0°C before you compute anything.

Example 30: Ice into a glass of water — does it all melt?

40 g of ice at 0°C is dropped into 250 g of water at 35°C in an insulated glass whose own heat capacity is negligible. Taking $s_{\text{water}} = 4186$ J/(kg K) and $L_f = 3.33 \times 10^{5}$ J/kg, find the final temperature.

Solution:

1. Do the branch test before anything else.

Heat available — what the warm water can give up before it reaches 0°C: $$Q_{\text{avail}} = m_{w}s_{w}\left(35 - 0\right) = 0.250 \times 4186 \times 35 = 3.663 \times 10^{4} \text{ J}$$

Heat required to melt all the ice: $$Q_{\text{req}} = m_{ice}L_f = 0.040 \times 3.33 \times 10^{5} = 1.332 \times 10^{4} \text{ J}$$

2. Compare. $3.663 \times 10^{4} > 1.332 \times 10^{4}$, so there is more than enough heat. Branch: all the ice melts, and the final temperature will be above 0°C. Now, and only now, write the balance.

3. The heat balance, with all three terms. The warm water cools from 35°C to $T$; the ice melts and then the meltwater warms from 0°C to $T$: $$m_{w}s_{w}\left(35 - T\right) = m_{ice}L_f + m_{ice}s_{w}\left(T - 0\right)$$

4. Substitute. $$0.250 \times 4186\left(35 - T\right) = 1.332\times 10^{4} + 0.040 \times 4186 \times T$$ $$1046.5\left(35 - T\right) = 13\,320 + 167.4\,T$$ $$36\,628 - 1046.5\,T = 13\,320 + 167.4\,T$$

5. Solve. $$23\,308 = 1213.9\,T \qquad \Longrightarrow \qquad T = 19.2°\text{C}$$

6. Check. The answer is above 0°C, which is consistent with the branch we chose — if it had come out negative, the branch would have been wrong and we would have had to go back. That self-consistency check is free and it catches the commonest error in the topic.

Final Answer: $19.2$°C. Branch taken: all 40 g of ice melts.

Takeaway: The branch test costs two multiplications and decides the whole answer. Do it before you write the balance, not after your answer comes out looking odd.

Example 31: The same question with far more ice

150 g of ice at 0°C is dropped into 120 g of water at 28°C, again in an insulated vessel of negligible heat capacity. Find the final temperature and the state of the contents.

Solution:

1. Branch test.

Heat available from the water cooling to 0°C: $$Q_{\text{avail}} = 0.120 \times 4186 \times 28 = 1.407 \times 10^{4} \text{ J}$$

Heat required to melt all 150 g of ice: $$Q_{\text{req}} = 0.150 \times 3.33 \times 10^{5} = 4.995 \times 10^{4} \text{ J}$$

2. Compare, and stop. $1.407 \times 10^{4} < 4.995 \times 10^{4}$ — the water can supply only about 28% of what is needed. Branch: partial melt. The melting stops when the water runs out of heat, and both the meltwater and the remaining ice sit at 0°C.

$$\boxed{T_{\text{final}} = 0°\text{C}}$$

3. How much ice actually melted. All the available heat went into melting: $$m_{\text{melted}} = \frac{Q_{\text{avail}}}{L_f} = \frac{1.407 \times 10^{4}}{3.33 \times 10^{5}} = 0.0422 \text{ kg} = 42.2 \text{ g}$$

4. The final contents.

• ice remaining: $150 - 42.2 = 107.8$ g at 0°C
• water: $120 + 42.2 = 162.2$ g at 0°C

3. The trap, spelled out. Anyone who wrote the full balance without testing first would have got $$0.120\times 4186\left(28 - T\right) = 0.150\times 3.33\times 10^{5} + 0.150 \times 4186\,T$$ which gives $T = -31.7$°C — a "final temperature" below the melting point of the ice that was supposed to have melted, and below the freezing point of water that was supposed to be liquid. Physically impossible. A negative answer here is the equation telling you that you chose the wrong branch.

Final Answer: $0$°C, with $107.8$ g of ice and $162.2$ g of water left in the vessel. Branch taken: partial melt.

Takeaway: When the answer pins at 0°C, the question stops being "what temperature?" and becomes "how much melted?" An impossible temperature is not a arithmetic slip — it is a branch error.

Example 32: Twenty grams of steam into a beaker

20 g of steam at 100°C is passed into 300 g of water at 20°C held in a copper calorimeter of water equivalent 0.020 kg. Taking $s_{\text{water}} = 4186$ J/(kg K) and $L_v = 22.6 \times 10^{5}$ J/kg, find the final temperature.

Solution:

1. Branch test, in the direction that matters here. The question is whether the steam can push everything up to 100°C — if it can, some steam is left uncondensed and the answer is exactly 100°C.

Heat available from condensing all the steam: $$Q_{\text{avail}} = m_{s}L_v = 0.020 \times 22.6 \times 10^{5} = 4.52 \times 10^{4} \text{ J}$$

Heat required to raise the water and calorimeter from 20°C to 100°C: $$Q_{\text{req}} = \left(0.300 + 0.020\right)\times 4186 \times 80 = 0.320 \times 4186 \times 80 = 1.072 \times 10^{5} \text{ J}$$

2. Compare. $4.52 \times 10^{4} < 1.072 \times 10^{5}$ — not nearly enough. Branch: all the steam condenses and the final temperature lies below 100°C.

3. Write the full balance. Steam condenses at 100°C, then the condensate cools from 100°C to $T$; the water and calorimeter warm from 20°C to $T$: $$m_{s}L_v + m_{s}s_{w}\left(100 - T\right) = \left(m_{w} + W\right)s_{w}\left(T - 20\right)$$

4. Substitute. $$4.52\times 10^{4} + 0.020 \times 4186\left(100 - T\right) = 0.320 \times 4186\left(T - 20\right)$$ $$45\,200 + 83.7\left(100 - T\right) = 1339.5\left(T - 20\right)$$ $$45\,200 + 8372 - 83.7\,T = 1339.5\,T - 26\,790$$

5. Solve. $$80\,362 = 1423.2\,T \qquad \Longrightarrow \qquad T = 56.5°\text{C}$$

6. Notice what did the work. Of the total $4.88 \times 10^{4}$ J the steam delivered, $4.52 \times 10^{4}$ J — 93% of it — came from condensation alone, before the condensate had cooled a single degree. Twenty grams of steam heated 320 grams of water by 36 K. This is exactly why steam scalds so much worse than boiling water, and why steam is used to heat buildings rather than hot water.

Final Answer: $56.5$°C. Branch taken: all the steam condenses.

Takeaway: When steam meets water, the latent heat usually dominates every other term. Compute $mL_v$ first — it will tell you at a glance whether the answer will be anywhere near 100°C.

Example 33: When the mixture ends up partly frozen

80 g of ice at $-20$°C is added to 60 g of water at 12°C in an insulated container. Taking $s_{\text{ice}} = 2100$ J/(kg K), $s_{\text{water}} = 4186$ J/(kg K) and $L_f = 3.33 \times 10^{5}$ J/kg, find the final temperature and the final contents.

Solution:

1. This one runs the other way, so test the other direction. Can the water supply enough heat merely to bring the ice up to 0°C? If not, the water will start to freeze.

Heat available from the water cooling from 12°C to 0°C: $$Q_{\text{avail}} = 0.060 \times 4186 \times 12 = 3014 \text{ J}$$

Heat required to warm the ice from $-20$°C to 0°C: $$Q_{\text{req}} = 0.080 \times 2100 \times 20 = 3360 \text{ J}$$

2. Compare. $3014 < 3360$ — the water cannot even get the ice to its melting point, let alone melt any of it. Branch: no ice melts; instead some of the water freezes, and the final temperature is 0°C.

3. Find the shortfall. $$3360 - 3014 = 346 \text{ J}$$ This must come from somewhere, and the only source left is the latent heat released when water at 0°C turns into ice at 0°C.

4. How much water freezes. $$m_{\text{frozen}} = \frac{346}{3.33 \times 10^{5}} = 1.04 \times 10^{-3} \text{ kg} = 1.04 \text{ g}$$

5. The final contents, at 0°C.

• ice: $80 + 1.04 = 81.0$ g
• water: $60 - 1.04 = 59.0$ g

2. A word on why this feels strange. Nothing melted; something froze. That is the correct physics whenever cold ice meets a small amount of barely-warm water — the ice is the bigger thermal reservoir here, and it wins. Notice that the answer was decided entirely by comparing 3014 with 3360; had the water started at 14°C instead of 12°C, the comparison would have flipped and some ice would have melted.

Final Answer: $0$°C, with $81.0$ g of ice and $59.0$ g of water. Branch taken: partial freezing of the water.

Takeaway: Cold ice can freeze the water you add to it. Test whether the warm body can even reach the melting point before you assume anything melts at all.