Quick Recap — Thermal Properties & Advanced Problems

  • Expansion: linear ΔL=αL ΔT\Delta L=\alpha L\,\Delta T; area β=2α\beta=2\alpha; volume γ=3α\gamma=3\alpha. Clamped rod thermal force =YAα ΔT=YA\alpha\,\Delta T.
  • Calorimetry: Q=mc ΔTQ=mc\,\Delta T (sensible), Q=mLQ=mL (latent); mixtures reach equilibrium by heat lost == heat gained. Water: Lf=80L_f=80 cal/g, Lv=540L_v=540 cal/g.
  • Conduction: Qt=kA ΔTL\dfrac{Q}{t}=\dfrac{kA\,\Delta T}{L}; thermal resistance LkA\dfrac{L}{kA} adds in series. Series junction of equal-AA, equal-LL rods: Tj=k1T1+k2T2k1+k2T_j=\dfrac{k_1T_1+k_2T_2}{k_1+k_2}.
  • Radiation: Stefan P∝AT4P\propto AT^4; Wien λmaxT=\lambda_{max}T= const; Newton's cooling: rate ∝(T−Ts)\propto(T-T_s).

Worked mini-example. 50 g of ice at 0∘0^\circC is added to 50 g of water at 40∘40^\circC. Heat available =50×40=2000=50\times40=2000 cal melts only 200080=25\dfrac{2000}{80}=25 g of ice, so equilibrium is at 0∘0^\circC with 2525 g of ice still unmelted.