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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2025 · 22 January · Morning Shift · Physics · Thermal Properties of Matter

An amount of ice of mass 10^ -3 ~kg and temperature -10^ C is...

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

Options

  1. A3043 J
  2. B3024 J
  3. C3003 J
  4. D3022 J

Correct answer

A. 3043 J

Step-by-step solution

An amount of ice of mass 10^ -3 ~kg and temperature -10^ C is transformed to vapour of temperature 110^ C by applying
ΔQ1=m×S1×ΔT=10−3×2100×10=21 JΔQ2=m×Lf=10−3×3.35×105=335 JΔQ3=m×Sw×ΔT=10−3×4180×100=418 JΔQ4=m×Lv=10−3×2.25×106=2250 JΔQ5=m×Sv×ΔT=10−3×1920×10=19.2 JΔQnet=3043.2 J\begin{aligned} & \Delta \mathrm{Q}_1=\mathrm{m} \times \mathrm{S}_1 \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}_{\mathrm{net}}=3043.2 \mathrm{~J}\end{aligned}
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