JEE Chemistry 2026 Question with Solution

Given: A cup of water is heated from $5^\circ \text{C}$ and begins to boil in a microwave oven.

Find: The correct signs of $q$, $w$, and $\Delta U$.

Using the first law of thermodynamics:

$$\Delta U = q + w$$

Energy is supplied to the water by the microwave, so heat enters the system. Therefore,

$$q > 0$$

As the water gets heated from $5^\circ \text{C}$ to the boiling stage, its internal energy increases. Therefore,

$$\Delta U > 0$$

When the water begins to boil, vapor forms and the system expands against atmospheric pressure. Hence work is done by the system on the surroundings. By the IUPAC convention,

$$w < 0$$

Therefore, the correct option is B: $q = +ve,\; w = -ve,\; \Delta U = +ve$.