Paper 2 Thermal Physics A Answers

Q1 Calorimetry

1. $Q_w=mc\Delta T=0.250(4200)(6.5)=6.83\times 10^3\mathrm{J}$. [2]
2. Metal temperature fall $=95.0-28.5=66.5\mathrm{K}$. $c=Q/(m\Delta T)=6825/(0.150\times 66.5)=684\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. [3]
3. Heat lost to surroundings, calorimeter heat capacity neglected, incomplete thermal equilibrium, or thermometer error. [1]

Q2 Heating curve

1. $E=Pt=45(160)=7.20\times 10^3\mathrm{J}$. [2]
2. $L=E/m=7200/0.080=9.0\times 10^4\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. [2]
3. Supplied energy changes particle arrangement/separation during melting rather than increasing average kinetic energy. [2]

Q3 Electrical specific heat capacity

1. $E=VIt=12.0(3.20)(240)=9.22\times 10^3\mathrm{J}$. [2]
2. $c=E/(m\Delta T)=9216/(0.520\times 15.0)=1.18\times 10^3\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. [2]
3. If heat loss is ignored, not all electrical energy heats the liquid, so calculated $c$ is too large. [2]