Q17 / 75 · JEE Main 2026 (23 January Shift 1) · Chemistry
A cup of water at $5^{\circ} \mathrm{C}$ (system) is placed in a microwave oven and the oven is turned on for one minute during which the water begins to boil. Which of the following option is true ?
- A$\mathrm{q}=+\mathrm{ve}, \mathrm{w}=0, \Delta \mathrm{U}=-\mathrm{ve}$
- B$\mathrm{q}=+\mathrm{ve}, \mathrm{w}=-\mathrm{ve}, \Delta \mathrm{U}=-\mathrm{ve}$
- C$\mathrm{q}=-\mathrm{ve}, \mathrm{w}=-\mathrm{ve}, \Delta \mathrm{U}=-\mathrm{ve}$
- Correct answerD$\mathrm{q}=+\mathrm{ve}, \mathrm{w}=-\mathrm{ve}, \Delta \mathrm{U}=+\mathrm{ve}$
Solution
For water heating in a microwave:
Heat (q): The microwave radiates electromagnetic energy that is absorbed by water molecules, so q = +ve
Work (w): As water heats and boils, it expands against atmospheric pressure. The system does work on the surroundings, so w = -ve
Internal Energy (ΔU): From the first law, ΔU = q + w. Since heat absorbed (+ve) is much larger in magnitude than work done against atmosphere (-ve), the net change is ΔU = +ve (confirmed by temperature increase from 5°C to boiling)
Therefore: q = +ve, w = -ve, ΔU = +ve
Heat (q): The microwave radiates electromagnetic energy that is absorbed by water molecules, so q = +ve
Work (w): As water heats and boils, it expands against atmospheric pressure. The system does work on the surroundings, so w = -ve
Internal Energy (ΔU): From the first law, ΔU = q + w. Since heat absorbed (+ve) is much larger in magnitude than work done against atmosphere (-ve), the net change is ΔU = +ve (confirmed by temperature increase from 5°C to boiling)
Therefore: q = +ve, w = -ve, ΔU = +ve
Hence, the correct answer is option: 'D'.