Question

Calculate the change in entropy when $1g$ of ice at $0^{o}C$ is heated to form water at ${40}^{o}C$

• A. $0.28cal{/}^{o}C$
• B. $1.411cal{/}^{o}C$
• C. $0.41cal{/}^{o}C$
• D. none of these

Answer 1

Answer: (C)

$0.41cal{/}^{o}C$

The change in entropy would be given as $\Delta S=\Delta S_1+\Delta S_2$ $\Delta S_1$ is to melt the ice and $\Delta S_2$ is to rise the temperature. Thus we get $\Delta S=\frac{mL}{T}+m{\int }_{T_1}^{T_2}\frac{dT}{T}$
$\Delta S=\frac{1\times 80}{273}+1\times$$\ln \frac{313}{273}=0.28+0.13=0.41cal{/}^{o}C$