Chapter 21, Problem 43AP

(a)

To determine

The entropy rise of the entire system.

Answer to Problem 43AP

The entropy rise of the entire system is $13.4\text{J}/\text{K}$.

Explanation of Solution

Given info: The mass of the athlete and the water is $70\text{kg}$ and $454\text{g}$ respectively. The initial temperature of athlete and water is $98.6°\text{F}$ and $35°\text{F}$ respectively.

Write the expression to calculate the change in entropy of the system.

$\Delta S=\Delta S_{\text{ice}\text{water}}+\Delta S_{\text{body}}$ (1)

Here,

$\Delta S_{\text{ice}\text{water}}$ is the change in the entropy of cold water.

$\Delta S_{\text{body}}$ is the change in the entropy of body.

$\Delta S$ is change in entropy of the system.

Write the expression to calculate the change in entropy of water.

$\Delta S_{\text{ice}\text{water}}=ms{\displaystyle \int \frac{dT}{T}}$ (2)

Here,

$m$ is the mass of water.

$s$ is the specific heat of water.

$T$ is the absolute temperature.

$dT$ is the change in temperature of water.

Write the expression to convert the temperature from Fahrenheit to Kelvin.

$\text{K}=\frac{5}{9}\left(°\text{F}-\text{32}\right)+273.15$ (3)

Substitute $98.6°\text{F}$ for $°\text{F}$ in equation (3).

$\begin{array}{c}\text{K}=\frac{5}{9}\left(98.6°\text{F}-\text{32}\right)+273.15\\ =310.15\text{K}\end{array}$

Thus, the temperature of body in Kelvin is $310.15\text{K}$.

Substitute $35°\text{F}$ for $°\text{F}$ in equation (3).

$\begin{array}{c}\text{K}=\frac{5}{9}\left(35°\text{F-32}\right)+273.15\\ =274.82\text{K}\end{array}$

Thus, the temperature of water in Kelvin is $274.82\text{K}$.

Substitute $454\text{g}$ for $m$, $4.18\text{J}/\text{g}\cdot \text{K}$ for $s$ in equation (1) to find $\Delta S_{\text{ice}\text{water}}$.

$\begin{array}{c}\Delta S_{\text{ice}\text{water}}=454\text{g}\left(4.18\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \int \frac{dT}{T}}\\ =1897.72\text{J}/\text{K}{\displaystyle \int \frac{dT}{T}}\end{array}$

Integrate the above expression from the limit of $274.67\text{K}$ to $310\text{K}$.

$\Delta S_{\text{ice}\text{water}}=1897.72\text{J}/\text{K}{\displaystyle \underset{274.67\text{K}}{\overset{310\text{K}}{\int }}\frac{dT}{T}}$

Write the expression to calculate the change in entropy of water.

$\Delta S_{\text{body}}=-\frac{ms\left(T_2-T_1\right)}{T_2}$

Here,

$T_2$ is the temperature of body.

$T_1$ is the temperature of water.

Substitute $-\frac{msdT}{T}$ for $\Delta S_{\text{body}}$ and $1897.72\text{J}/\text{K}{\displaystyle \underset{274.67\text{K}}{\overset{310\text{K}}{\int }}\frac{dT}{T}}$ for $\Delta S_{\text{ice}\text{water}}$ in equation (1).

$\Delta S=1897.72\text{J}/\text{K}{\displaystyle \underset{274.67\text{K}}{\overset{310\text{K}}{\int }}\frac{dT}{T}}+-\frac{ms\left(T_2-T_1\right)}{T_2}$                               (4)

Substitute $454\text{g}$ for $m$, $4.18\text{J}/\text{g}\cdot \text{K}$ for $s$, $310\text{K}$ for $T_2$ and $274.67\text{K}$ for $T_1$ in equation (4) to find $\Delta S$.

$\begin{array}{c}\Delta S=1897.72\text{J}/\text{K}{\displaystyle \underset{274.67\text{K}}{\overset{310\text{K}}{\int }}\frac{dT}{T}}-\frac{454\text{g}\times 4.18\text{J}/\text{g}\cdot \text{K}\left(310\text{K}-274.67\text{K}\right)}{310\text{K}}\\ =1897.72\text{J}/\text{K}\times \ln \left(\frac{310\text{K}}{274.67\text{K}}\right)-216.27\text{J}/\text{K}\\ =13.4\text{J}/\text{K}\end{array}$

Thus, the entropy rise of the entire system is $13.4\text{J}/\text{K}$.

Conclusion:

Therefore, the entropy rise of the entire system is $13.4\text{J}/\text{K}$.

(b)

To determine

The athlete’s temperature after she drinks the cold water.

Answer to Problem 43AP

The final temperature of the body is $310\text{K}$.

Explanation of Solution

Given info: The mass of the athlete and the water is $70\text{kg}$ and $454\text{g}$ respectively. The initial temperature of athlete and water is $98.6°\text{F}$ and $35°\text{F}$ respectively.

Write the expression of heat balance equation.

$\begin{array}{c}\text{Heat}\text{gained}\text{by}\text{water}=\text{Heat}\text{lost}\text{by}\text{body}\\ ms\left(T_{\text{f}}-274.82\text{K}\right)=Ms\left(310.15\text{K}-{\text{T}}_{\text{f}}\right)\\ m\left(T_{\text{f}}-274.82\text{K}\right)=M\left(310.15\text{K}-{\text{T}}_{\text{f}}\right)\end{array}$ (5)

Here,

$T_{\text{f}}$ is the final temperature of the body.

Substitute $454\text{g}$ for $m$ and $70\text{kg}$ for $M$ in equation (5).

$\begin{array}{c}454\text{g}\left(T_{\text{f}}-274.82\text{K}\right)=70\text{kg}\times \frac{1000\text{g}}{1\text{kg}}\left(310.15{\text{K-T}}_{\text{f}}\right)\\ T_{\text{f}}=309.92\text{K}\\ \approx \text{310}\text{K}\end{array}$

Conclusion:

Therefore, the final temperature of the body is $310\text{K}$.

(c)

To determine

The entropy rise of the entire system.

Answer to Problem 43AP

The entropy rise of the entire system is $13.3\text{J}/\text{K}$.

Explanation of Solution

Given info: The mass of the athlete and the water is $70\text{kg}$ and $454\text{g}$ respectively. The initial temperature of athlete and water is $98.6°\text{F}$ and $35°\text{F}$ respectively.

Write the expression to calculate the change in entropy of the system.

$\Delta S=\Delta S_{\text{ice}\text{water}}+\Delta S_{\text{body}}$ (1)

Write the expression to calculate the change in entropy of water.

$\Delta S_{\text{ice}\text{water}}=ms{\displaystyle \int \frac{dT}{T}}$ (2)

Integrate the above expression from the limit of $274.67\text{K}$ to $309.78\text{K}$.

$\Delta S_{\text{ice}\text{water}}=ms{\displaystyle \underset{274.82\text{K}}{\overset{309.92\text{K}}{\int }}\frac{dT}{T}}$

Substitute $454\text{g}$ for $m$, $\left(1\text{cal}/\text{g}\cdot \text{K}\right)$ for $s$ in above equation.

$\begin{array}{c}\Delta S_{\text{ice}\text{water}}=454\text{g}\left(1\text{cal}/\text{g}\cdot \text{K}\right){\displaystyle \underset{274.82\text{K}}{\overset{309.92\text{K}}{\int }}\frac{dT}{T}}\\ =54.6\text{cal}/\text{K}\\ \approx 55\text{cal}/\text{K}\end{array}$

Write the expression to calculate the change in entropy of body.

$\Delta S_{\text{body}}=Ms{\displaystyle \int \frac{dT}{T}}$ (7)

Here,

$M$ is the mass of athlete.

Integrate the above expression from the limit of $309.92\text{K}$ to $310.15\text{K}$.

$\Delta S_{\text{body}}=Ms{\displaystyle \underset{310.15\text{K}}{\overset{309.92\text{K}}{\int }}\frac{dT}{T}}$

Substitute $70\text{kg}$ for $M$ and $\left(1\text{cal}/\text{g}\cdot \text{K}\right)$ for $s$ in above equation.

$\begin{array}{c}\Delta S_{\text{body}}=70\text{kg}\times \frac{1000\text{g}}{1\text{kg}}\left(1\text{cal}/\text{g}\cdot \text{K}\right){\displaystyle \underset{310.15\text{K}}{\overset{309.92\text{K}}{\int }}\frac{dT}{T}}\\ =-51.8\text{cal}/\text{K}\end{array}$

Substitute $55\text{cal}/\text{K}$ for $\Delta S_{\text{ice}\text{water}}$ and $-51.8\text{cal}/\text{K}$ for $\Delta S_{\text{body}}$ in equation (1) to find $\Delta S$.

` $\begin{array}{c}\Delta S=55\text{cal}/\text{K}-51.8\text{cal}/\text{K}\\ =3.2\text{cal}/\text{K}\\ =3.2\text{cal}/\text{K}\times \frac{4.18\text{J}}{1\text{cal}}\\ \approx 13.3\text{J}/\text{K}\end{array}$ (8)

Thus, the entropy rise of the entire system is $13.3\text{J}/\text{K}$.

Conclusion:

Therefore, the entropy rise of the entire system is $13.3\text{J}/\text{K}$.

(d)

To determine

The result by comparing the part (a) and (c).

Answer to Problem 43AP

The change in entropy in part (c) is less than that of part (a) by less than 1%.

Explanation of Solution

Given info: The mass of the athlete and the water is $70\text{kg}$ and $454\text{g}$ respectively. The initial temperature of athlete and water is $98.6°\text{F}$ and $35°\text{F}$ respectively.

The percentage change in entropy is,

$\begin{array}{c}\%S=\frac{13.4\text{J}/\text{K}-13.3\text{J}/\text{K}}{13.4\text{J}/\text{K}}\times 100\\ =0.74\%\end{array}$

Thus the change in entropy in part (c) is less than that of part (a) by less than 1%.

Conclusion:

Therefore, the change in entropy in part (c) is less than that of part (a) by less than 1%.