Thermal Properties of Matter

JEE Main level, single correct, medium

56 of 126

Question 56 of 126

1kg1kg of water at 100∘C100^{\circ}C is converted into steam at 100∘C100^{\circ}C by boiling at atmospheric pressure. The volume of water changes from 1.00×10−3m31.00\times {10}^{-3}{m}^{3} as a liquid to 1.671m31.671{m}^{3} as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisation =2257kJ/kg=2257kJ/kg, Atmospheric pressure =1×105Pa)\left.=1\times {10}^{5}Pa\right)

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Answer: (B) +2090kJ+2090kJ

The work to be done in the process is given by
dW=PdV=1×105Pa×(1.671−0.001)m3=1.670×105J\begin{matrix}dW & = & PdV \\ & = & 1\times {10}^{5}Pa\times (1.671-0.001){m}^{3} \\ & = & 1.670\times {10}^{5}J\end{matrix}
The change in heat energy during the vaporisation process can be calculated as follows-
ΔQsupplied=2257×1×103J=22.57×105J\begin{matrix}\Delta {Q}_{\text{supplied}} & = & 2257\times 1\times {10}^{3}J \\ & = & 22.57\times {10}^{5}J\end{matrix}
Hence, the change in internal energy in the process is given by
ΔU=ΔQsupplied−ΔW=(22.57−1.67)×105J=20.9×105J=2090kJ\begin{matrix}\Delta U & = & \Delta {Q}_{supplied}-\Delta W \\ & = & (22.57-1.67)\times {10}^{5}J \\ & = & 20.9\times {10}^{5}J \\ & = & 2090kJ\end{matrix}

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