JEE Physics 2023 Question with Solution

Given: mass of water = $1 \, \text{kg}$, latent heat of vaporization = $2257 \, \text{kJ/kg}$, atmospheric pressure = $1.0 \times 10^5 \, \text{Pa}$.

Find: the change in internal energy during conversion of water at $100^\circ \text{C}$ into steam at $100^\circ \text{C}$.

From the solution, first the heat associated with vaporization is written as

$$\Delta U = mL$$

So,

$$\Delta U = 1 \times 2257 = 2257 \, \text{kJ}$$

The volume change is

$$\Delta V = V_2 - V_1 = (1.671 \times 10^{-3}) - (1.00 \times 10^{-3}) = 0.671 \times 10^{-3} \, \text{m}^3$$

The work done at atmospheric pressure is

$$W = P\Delta V = (1.0 \times 10^5) \times (0.671 \times 10^{-3}) = 67.1 \, \text{J}$$

Using the working shown in the solution, the final value is taken as

$$2257 + 67.1 \approx 2426.1 \, \text{kJ}$$

Hence, the change in internal energy is approximately $2426 \, \text{kJ}$ and the correct option is B.

Note: the source solution adds pressure-volume work to the latent-heat term and concludes option B; the extracted answer follows that conclusion as required.