NEET Physics Question - Thermal Physics

A sample of $0.1 \text{ g}$ of water at $100^{\circ}\text{C}$ and normal pressure ($1.013 \times 10^{5} \text{ Nm}^{-2}$) requires $54 \text{ cal}$ of heat energy to convert to steam at $100^{\circ}\text{C}$. If the volume of the steam produced is $167.1 \text{ cc}$, the change in internal energy of the sample, is: (2018)

$42.2 \text{ J}$

$208.7 \text{ J}$

$104.3 \text{ J}$

$84.5 \text{ J}$

Solution:

Heat supplied $Q = 54 \text{ cal} = 54 \times 4.18 \text{ J} = 225.72 \text{ J}$. Work done $W = P\Delta V = 1.013 \times 10^{5} \times (167.1 - 0.1) \times 10^{-6} \approx 16.92 \text{ J}$. By first law of thermodynamics, $\Delta U = Q - W = 225.72 - 16.92 = 208.8 \text{ J}$, which is closest to $208.7 \text{ J}$.

Rankers Physics

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