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Q. The average molar heat capacities of ice and water are $37.6$ and $75.2\, J\, mol ^{-1} K ^{-1}$ respectively and its enthalpy of fusion is $6.02\, kJ\, mol ^{-1} .$ The amount of heat required to raise the temperature of $10\, g$ of water from $-10^{\circ} C$ to $10^{\circ} C$ is equal to

Thermodynamics

Solution:

$10 g$ ice at $-10^{\circ} C \stackrel{ I }{\longrightarrow } 10 g$ ice at $0^{\circ} C \stackrel{ II }{\longrightarrow }$

$10 g$ water at $0^{\circ} C \stackrel{ III }{\longrightarrow } 10 g$ water at $10^{\circ} C$

Heat required $=\frac{37.6}{18}\underset{I}{\times 10 \times 10}+ \underset{II}{\frac{6020 \times 10}{18}}+\frac{75.2}{18}\underset{III}{ \times 10 \times 10}$

$=209+3344+418=3971\, J$