Chapter 15, Problem 38P

To determine

To find: The change in entropy

Answer to Problem 38P

  $\text{[Δ}\mathrm{S}=11\text{kcal}/\text{K}]=5\times {10}^4\text{J}/\text{K}$

Explanation of Solution

Given:

  $1.00{\text{m}}^3$ of water at $0^{\circ }\text{C}$ is frozen

It is cooled to $-{10}^{\circ }\text{C}$ by being in contact with an ice at $-{10}^{\circ }\text{C}$

Formula used:

  $\text{Δ}{\mathrm{S}}_{\mathrm{w}}=-\frac{\mathrm{Q}}{\mathrm{T}}$

  $\text{Δ}{\mathrm{S}}_{\mathrm{i}}=\frac{\mathrm{Q}}{\mathrm{T}}$

Calculation:

Volume of water $(\text{V})=1{\text{m}}^3$

Mass of water $(\mathrm{m})=1{\text{m}}^3\times {10}^3\text{kg}/{\text{m}}^3={10}^3\text{kg}$

Temperature of ice $(\mathrm{T})=-{10}^{\circ }\text{C}$

Latent heat of fusion of ice $\left({\mathrm{L}}_{\mathrm{f}}\right)=79.7\text{kcals}/\text{kg}$

First determine the mass of ice

Heat lost by water = heat gained by ice

  ${\mathrm{m}}_{\mathrm{i}\mathrm{r}}{\mathrm{L}}_{\mathrm{f}}={\mathrm{m}}_{\mathrm{i}}\times {\mathrm{C}}_{\mathrm{i}}\times \text{Δ}\mathrm{T}$

  ${\mathrm{m}}_{\mathrm{i}}=\frac{{\mathrm{m}}_{\mathrm{u}}{\mathrm{L}}_{\mathrm{f}}}{{\mathrm{C}}_{\mathrm{i}}\times \text{Δ}\mathrm{T}}$

  $=\frac{{10}^3\times 79.7\times {10}^8}{0.50\times {10}^8[0-(-10)]}\text{kg}$

  $=1.594\times {10}^4\text{kg}$

Change in entropy of water is,

  $\begin{array}{c}\text{Δ}{\mathrm{S}}_{\mathrm{w}}=-\frac{\mathrm{Q}}{\mathrm{T}}\\ \text{Δ}{\mathrm{S}}_{\mathrm{w}}=-\frac{{10}^3\times 79.7\text{kcal}}{273}\\ =-291.9\text{kcals}/\text{K}\end{array}$

Change in entropy of ice is,

  $\begin{array}{c}\text{Δ}{\mathrm{S}}_{\mathrm{i}}=\frac{\mathrm{Q}}{\mathrm{T}}\\ =\frac{1.594\times {10}^4\times 0.5\times 10}{263}\\ \text{Δ}{\mathrm{S}}_{\mathrm{i}}=303\text{kcals}/\text{K}\end{array}$

Total change in entropy is,

  $\begin{array}{c}\text{ Δ}\mathrm{S}=\text{Δ}{\mathrm{S}}_{\mathrm{w}}+{\mathrm{S}}_{\mathrm{i}}\\ =-291.9\text{kcal}/\mathrm{K}+303\text{kcal}/\mathrm{K}\\ =11.1\text{kcal}/\text{K}\\ \approx 5\times {10}^4\text{J}/\text{K}\end{array}$

Conclusion:

Therefore, change in entropy is $\text{[Δ}\mathrm{S}=11\text{kcal}/\text{K}]=5\times {10}^4\text{J}/\text{K}$