Physics · Thermodynamics

JEE Main 2025 — 22 January, Morning Shift — Question 50

An amount of ice of mass $10^{-3} \mathrm{~kg}$ and temperature

$-10^{\circ} \mathrm{C}$ is transformed to vapour of temperature $110^{\circ}$ by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice $=2100 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ , specific heat of water $=4180 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ , specific heat of steam $=1920 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ , Latent heat of ice $=3.35 \times 10^{5} \mathrm{Jkg}^{-1}$ and Latent heat of steam $=2.25 \times 10^{6} \mathrm{Jkg}^{-1}$ )

Option A:
3022 J
Option B:
3043 J
Correct
Option C:
3003 J
Option D:
3024 J

Answer: B

Step-by-step solution

$\Delta \mathrm{Q}_{1}=\mathrm{m} \times \mathrm{S}_{\mathrm{I}} \times \Delta \mathrm{T}=10^{-3} \times 2100 \times 10=21 \mathrm{~J}$

$\Delta \mathrm{Q}_{2}=\mathrm{m} \times \mathrm{L}_{\mathrm{f}}=10^{-3} \times 3.35 \times 10^{5}=335 \mathrm{~J}$

$\Delta \mathrm{Q}_{3}=\mathrm{m} \times \mathrm{S}_{\mathrm{w}} \times \Delta \mathrm{T}=10^{-3} \times 4180 \times 100=418 \mathrm{~J}$

$\Delta \mathrm{Q}_{4}=\mathrm{m} \times \mathrm{L}_{\mathrm{v}}=10^{-3} \times 2.25 \times 10^{6}=2250 \mathrm{~J}$

$\Delta \mathrm{Q}_{5}=\mathrm{m} \times \mathrm{S}_{\mathrm{v}} \times \Delta \mathrm{T}=10^{-3} \times 1920 \times 10=19.2 \mathrm{~J}$

$\Delta \mathrm{Q}_{\text {net }}=3043.2 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy