Question 168: Thermodynamics & KTG

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 $=10^5 \mathrm{~Pa}$ )

Accepted Answer

Heat supplied to water:
$$Q=mc\Delta T$$
Here, $m=4\,	ext{kg}$, $c=4.2	imes 10^3\,	ext{J kg}^{-1}	ext{K}^{-1}$, and $\Delta T = 20-4 = 16\,	ext{K}$.
So,
$$Q=4	imes 4.2	imes 10^3 	imes 16 = 268800\,	ext{J}$$
Since heating is done at atmospheric pressure, water expands and does work:
$$W = P\Delta V$$
Initial volume:
$$V_1=\frac{m}{ho_1}=\frac{4}{1000}=0.004\,	ext{m}^3$$
Final volume:
$$V_2=\frac{m}{ho_2}=\frac{4}{998}=0.004008016\,	ext{m}^3$$
Change in volume:
$$\Delta V=V_2-V_1=0.004008016-0.004=8.016	imes 10^{-6}\,	ext{m}^3$$
Work done:
$$W = 10^5 	imes 8.016	imes 10^{-6} = 0.8016\,	ext{J}$$
By first law of thermodynamics,
$$\Delta U = Q - W = 268800 - 0.8016 = 268799.1984\,	ext{J} \approx 268799.2\,	ext{J}$$
Correct option: A (268799.2 J)

Answer

268799.2

Answer

315826.2

Answer

234699.2

Answer

258700.8

Thermodynamics & KTG

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