Verified Solution • 2025–26 Session moderate

QJEEM25Pha748: JEE Main 2025 Physics - Class 11 JEEM25Pha748 Solution: An amount of ice of mass and temperature is transformed to…

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} )

Multiple Choice Options

  • A 3022 J
  • B 3043 J
  • C 3003 J
  • D 3024 J

Step-by-Step Verified Solution

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Heat and Thermodynamics Question 20 English Explanation

amp;ΔQ1=m×S1×ΔT=10−3×2100×10=21 Jamp;ΔQ2=m×Lf=10−3×3.35×105=335 Jamp;ΔQ3=m×Sw×ΔT=10−3×4180×100=418 Jamp;ΔQ4=m×Lv=10−3×2.25×106=2250 Jamp;ΔQ5=m×Sv×ΔT=10−3×1920×10=19.2 Jamp;ΔQnet =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}_{\text {net }}=3043.2 \mathrm{~J}\end{aligned}

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