Chapter 15, Problem 35P

To determine

To find: the change in entropy

Answer to Problem 35P

The change in entropy is $1.5\times {10}^3\text{J}/\text{K}$

Explanation of Solution

Given:

  $250$ gram of steam condensed to water $\mathrm{T}=100\text{°C}$

Formula used:

The expression of change in entropy,

  $\begin{array}{l}\text{Δ}\mathrm{S}=\int \mathrm{d}\mathrm{S}\hfill \\ \text{And, }\mathrm{d}\mathrm{S}=\frac{\mathrm{\delta }\mathrm{q}}{\mathrm{T}}\hfill \end{array}$

Then,

  $\begin{array}{l}\text{Δ}\mathrm{S}=\int \frac{\mathrm{\delta }\mathrm{q}}{\mathrm{T}}\\ =\frac{\mathrm{Q}}{\mathrm{T}}\end{array}$

Here $\mathrm{Q}$ is heat energy and $\mathrm{T}$ is temperature.

Heat energy for evaporation

  $\mathrm{Q}=-\mathrm{m}{\mathrm{L}}_{\text{vap}}$

Here $\mathrm{m}$ is mass and ${\mathrm{L}}_{\text{vap}}$ is latent heat of evaporation.

Then change in entropy is,

  $\text{Δ}\mathrm{S}=-\frac{\mathrm{m}{\mathrm{L}}_{\text{ve}}}{\mathrm{T}}$

Calculation:

Now substitute $250\text{g}$ for $\mathrm{m},22.6\times {10}^5\text{J}/\text{kg}$ for ${\mathrm{L}}_{\text{vap}}$ and ${100}^{\circ }\text{C}$ for $\mathrm{T}$ in entropy change equation.

  $\begin{array}{c}\text{Δ}\mathrm{S}=-\frac{(250\text{g})\left(22.6\times {10}^5\text{J}/\text{kg}\right)}{{100}^{\circ }\text{C}}\\ =-\frac{(250\text{g})\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(22.6\times {10}^5\text{J}/\text{kg}\right)}{(100+273)\text{K}}\\ =-\frac{\left(250\times {10}^{-3}\text{kg}\right)\left(22.6\times {10}^5\text{J}/\text{kg}\right)}{(373)\text{K}}\\ =-1514.75\text{J}/\text{K}\end{array}$

Conclusion:

Hence the change in entropy is $1.5\times {10}^3\text{J}/\text{K}$