Question 54 of 126

An amount of ice of mass $10^{-3} \mathrm{~kg}$ and temperature $-10^{\circ} \mathrm{C}$ is transformed to vapour of temperature $110^{\circ} \mathrm{C}$ 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 $=2.25 \times 10^6$ $\mathrm{Jkg}^{-1}$ )

A3043 JB3024 JC3003 JD3022 J

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Answer: (A) 3043 J

$\begin{aligned} & \Delta \mathrm{Q}_1=\mathrm{m} \times \mathrm{S}_1 \times \Delta \mathrm{T}=10^{-3} \times 2100 \times 10=21 \mathrm{~J} \\ & \Delta \mathrm{Q}_2=\mathrm{m} \times \mathrm{L}_{\mathrm{f}}=10^{-3} \times 3.35 \times 10^5=335 \mathrm{~J} \\ & \Delta \mathrm{Q}_3=\mathrm{m} \times \mathrm{S}_{\mathrm{w}} \times \Delta \mathrm{T}=10^{-3} \times 4180 \times 100=418 \mathrm{~J} \\ & \Delta \mathrm{Q}_4=\mathrm{m} \times \mathrm{L}_{\mathrm{v}}=10^{-3} \times 2.25 \times 10^6=2250 \mathrm{~J} \\ & \Delta \mathrm{Q}_5=\mathrm{m} \times \mathrm{S}_{\mathrm{v}} \times \Delta \mathrm{T}=10^{-3} \times 1920 \times 10=19.2 \mathrm{~J} \\ & \Delta \mathrm{Q}_{\mathrm{net}}=3043.2 \mathrm{~J}\end{aligned}$