Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 23 January, Morning Shift — Question 55

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?

Option A:
$q=+v e, w=0, \Delta U=-v e$
Option B:
$q=+v e, w=-v e, \Delta U=+v e$
Correct
Option C:
$q=-v e, w=-v e, \Delta U=-v e$
Option D:
$q=+v e, w=-v e, \Delta U=-v e$

Answer: B

Step-by-step solution

$\underset{5^{\circ} \mathrm{C}}{\mathrm{H}_{2} \mathrm{O}(\ell)} \longrightarrow \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_{2} \mathrm{O}(\ell)} \rightleftarrows \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_{2} \mathrm{O}(\mathrm{g})}$ due to expansion $\mathrm{w}=-\mathrm{ve}$ as heat is given to system so $\mathrm{q}=+\mathrm{ve}$ and internal energy of gas will be more than internal energy of liquid so $\Delta \mathrm{U}=+\mathrm{ve}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Introduction to Terminology in Thermodynamics