Density of water at $4^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$ are $1000 \mathrm{~kg} / \mat

Question

Density of water at $4^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$ are $1000 \mathrm{~kg} / \mathrm{m}^3$ and $998 \mathrm{~kg} / \mathrm{m}^3$ respectively. The increase in internal energy of 4 kg of water when it is heated from $4{ }^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$ is $\_\_\_\_$ J.
(specific heat capacity of water $=4.2 \mathrm{~J} / \mathrm{kg}$. and 1 atmospheric pressure $=10^5 \mathrm{~Pa}$ )

Professor PC Thomas Classes · Answer

The heat added to the water is $Q = mc\Delta T$. Assuming the specific heat given is $4.2 	ext{ kJ/kg}\cdot^\circ	ext{C}$ (standard value $4200 	ext{ J/kg}\cdot^\circ	ext{C}$), $Q = 4 	imes 4200 	imes (20 - 4) = 16 	imes 4200 = 268800 	ext{ J}$. The work done by expansion against atmospheric pressure is $W = P\Delta V = P(\frac{m}{ho_2} - \frac{m}{ho_1}) = 10^5 	imes 4 	imes (\frac{1}{998} - \frac{1}{1000}) = 4 	imes 10^5 	imes (\frac{2}{998000}) \approx 0.8 	ext{ J}$. By the first law of thermodynamics, the change in internal energy is $\Delta U = Q - W = 268800 - 0.8 = 268799.2 	ext{ J}$.