AIIMS 2018 Physics First Law of Thermodynamics MCQ Question

To calculate the increase in internal energy when 1 cm³ of H₂O is vaporized, we first determine the heat absorbed during vaporization using the latent heat of vaporization: $Q = m \cdot L$, where $m = 1 \, \text{g}$ (since the density of water is $1 \, \text{g/cm}^3$) and $L = 540 \, \text{cal/g}$. This gives $Q = 1 \, \text{g} \times 540 \, \text{cal/g} = 540 \, \text{cal}$. The work done by the steam as it expands to 1671 cm³ at 1 atm is calculated using $W = P \Delta V$, where $P = 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa}$ and $\Delta V = 1671 \, \text{cm}^3 = 1.671 \times 10^{-3} \, \text{m}^3$, yielding $W \approx 173 \, \text{cal}$. The increase in internal energy is then $ \Delta U = Q - W = 540 , \text{cal} - 173 , \text{cal} = 367 , \