MHTCET 16 Apr 2026 Shift 2 Paper

Concept:The heat supplied during vaporisation is utilised as external work done and increase in internal energy, as stated by the first law of thermodynamics.Explanation:For this phase change from liquid to vapour, the first law of thermodynamics gives:$$Q = \Delta U + W$$where $Q$ is the total heat supplied, $\Delta U$ is the increase in internal energy, and $W$ is the work done during expansion.The work done when the volume increases at constant pressure is:$$W = P \Delta V$$Given, $P = 3 \times 10^{5}\ \text{Pa}$ and $\Delta V = 1600\ \text{cm}^3$.Convert the volume into SI units:$$\Delta V = 1600 \times 10^{-6}\ \text{m}^3 = 1.6 \times 10^{-3}\ \text{m}^3$$Thus, the work done is:$$W = 3 \times 10^{5} \times 1.6 \times 10^{-3} = 480\ \text{J}$$According to the question, this work is $10\%$ of the heat supplied:$$480 = \frac{10}{100}Q$$$$Q = 4800\ \text{J}$$Using the first law of thermodynamics again:$$\Delta U = Q - W = 4800 - 480$$$$\Delta U = 4320\ \text{J}$$Answer:The increase in internal energy during the process is $4320\ \text{J}$.Correct option is C.