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An amount of ice of mass 10^-3 ~kg and temperature -10° C is transformed to vapour of temperature 110° C by applying heat. The total amount of work…

JEE Main 2025 — Physics Thermodynamics

2025mcqmedium

An amount of ice of mass 10−3 kg10^{-3} \mathrm{~kg}10−3 kg and temperature −10∘C-10^{\circ} \mathrm{C}−10∘C is transformed to vapour of temperature 110∘C110^{\circ} \mathrm{C}110∘C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice =2100Jkg−1 K−1=2100 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}=2100Jkg−1 K−1, specific heat of water =4180Jkg−1 K−1=4180 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}=4180Jkg−1 K−1, specific heat of steam =1920Jkg−1 K−1=1920 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}=1920Jkg−1 K−1, Latent heat of ice =3.35×105Jkg−1=3.35 \times 10^5 \mathrm{Jkg}^{-1}=3.35×105Jkg−1 and Latent heat of steam =2.25×106=2.25 \times 10^6=2.25×106 Jkg−1\mathrm{Jkg}^{-1}Jkg−1 )

Official previous-year question

Held on 22 Jan 2025 · Verified 6 Jul 2026.

Options

  1. A

    3043 J

  2. B

    3024 J

  3. C

    3003 J

  4. D

    3022 J

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