Chapter 22, Problem 22.78AP

An athlete whose mass is 70.0 kg drinks 16.0 ounces (454 g) of refrigerated water. The water is at a temperature of 35.0°F. (a) Ignoring the temperature change of the body that results from the water intake (so that the body is regarded as a reservoir always at 98.6°F), find the entropy increase of the entire system. (b) What If? Assume the entire body is cooled by the drink and the average specific heat of a person is equal to the specific heat of liquid water. Ignoring any other energy transfers by heat and any metabolic energy release, find the athlete’s temperature after she drinks the cold water given an initial body temperature of 98.6°F. (c) Under these assumptions, what is the entropy increase of the entire system? (d) State how this result compares with the one you obtained in part (a).

To determine

The entropy rise of the entire system.

Answer to Problem 22.78AP

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

Explanation of Solution

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.

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

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

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

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

Here, $m$ is the mass of water, $s$ is the specific heat of water, $T$ is the absolute temperature and $dT$ is the change in temperature of water.

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

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

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

    $\begin{array}{c}{\text{T}}_{\text{2}}=\frac{5}{9}\left(98.6°\text{F}-\text{32}\right)+273.15\\ =310.15\text{K}\end{array}$

Here, $T_2$ is the temperature of body.

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

    $\begin{array}{c}{\text{T}}_{\text{1}}=\frac{5}{9}\left(35°\text{F-32}\right)+273.15\\ =274.82\text{K}\end{array}$

Here, $T_1$ is the temperature of water.

Substitute $454\text{g}$ for $m$, $4.186\text{J}/\text{g}\cdot \text{K}$ for $s$ in equation (I) to find $\Delta S_{\text{ice}\text{water}}$.

    $\Delta S_{\text{ice}\text{water}}=454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \int \frac{dT}{T}}$

Integrate the above expression from the limit of $274.82\text{K}$ to $310.15\text{K}$.

    $\Delta S_{\text{ice}\text{water}}=454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \underset{274.82\text{K}}{\overset{310.15\text{K}}{\int }}\frac{dT}{T}}$

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

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

Substitute $-\frac{ms\left(T_2-T_1\right)}{T_2}$ for $\Delta S_{\text{body}}$ and $454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \underset{274.82\text{K}}{\overset{310.15\text{K}}{\int }}\frac{dT}{T}}$ for $\Delta S_{\text{ice}\text{water}}$ in equation (I).

    $\Delta S=454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \underset{274.82\text{K}}{\overset{310.15\text{K}}{\int }}\frac{dT}{T}}+-\frac{ms\left(T_2-T_1\right)}{T_2}$                                (IV)

Conclusion:

Substitute $454\text{g}$ for $m$, $4.186\text{J}/\text{g}\cdot \text{K}$ for $s$, $310.15\text{K}$ for $T_2$ and $274.82\text{K}$ for $T_1$ in equation (V) to find $\Delta S$.

    $\begin{array}{c}\Delta S=454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \underset{274.82\text{K}}{\overset{310.15\text{K}}{\int }}\frac{dT}{T}}-\frac{454\text{g}\times 4.186\text{J}/\text{g}\cdot \text{K}\left(310.15\text{K}-274.82\text{K}\right)}{310.15\text{K}}\\ =454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right)\times \ln \left(\frac{310.15\text{K}}{274.82\text{K}}\right)-216.48\text{J}/\text{K}\\ =13.4\text{J}/\text{K}\end{array}$

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

To determine

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

Answer to Problem 22.78AP

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

Explanation of Solution

Write the expression of heat balance equation.

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

Here, $T_{\text{f}}$ is the final temperature of the body, $m$ is the mass of the water and $M$ is the mass of the athlete.

Conclusion:

Substitute $454\text{g}$ for $m$ and $70\text{kg}$ for $M$ in equation (V) to find $T_{\text{f}}$.

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

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

To determine

The entropy rise of the entire system.

Answer to Problem 22.78AP

The entropy rise of the entire system is $11.1\text{J}/\text{K}$.

Explanation of Solution

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.

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

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

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

Integrate the above expression from the limit of $274.82\text{K}$ 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 $m$, $4.186\text{J}/\text{g}\cdot \text{K}$ for $s$ in above equation.

    $\begin{array}{c}\Delta S_{\text{ice}\text{water}}=454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \underset{274.82\text{K}}{\overset{309.92\text{K}}{\int }}\frac{dT}{T}}\\ =454\text{g}\left(4.186\text{J}/\text{g}\cdot \text{K}\right)\ln \left(\frac{309.92\text{K}}{274.82\text{K}}\right)\\ =229.84\text{J}/\text{K}\end{array}$

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

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

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 $4.186\text{J}/\text{g}\cdot \text{K}$ for $s$ in above equation.

    $\begin{array}{c}\Delta S_{\text{body}}=70\text{kg}\times \frac{1000\text{g}}{1\text{kg}}\left(4.186\text{J}/\text{g}\cdot \text{K}\right){\displaystyle \underset{310.15\text{K}}{\overset{309.92\text{K}}{\int }}\frac{dT}{T}}\\ =70\text{kg}\times \frac{1000\text{g}}{1\text{kg}}\left(4.186\text{J}/\text{g}\cdot \text{K}\right)\ln \left(\frac{309.92\text{K}}{310.15}\right)\\ =-217.38\text{J}/\text{K}\end{array}$

Conclusion:

Substitute $229.84\text{J}/\text{K}$ for $\Delta S_{\text{ice}\text{water}}$ and $-217.38\text{J}/\text{K}$ for $\Delta S_{\text{body}}$ in equation (III) to find $\Delta S$.

`    $\begin{array}{c}\Delta S=228.43\text{J}/\text{K}-217.38\text{J}/\text{K}\\ =11.05\text{J}/\text{K}\\ \approx 11.1\text{J}/\text{K}\end{array}$                           (VIII)

Thus, the entropy rise of the entire system is $11.1\text{J}/\text{K}$.

To determine

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

Answer to Problem 22.78AP

The the change in entropy in part (c) is less than that of part (a) by a factor of $0.828$.

Explanation of Solution

Write the expression for the ratio of entropy in part (c) and (a)

    $\begin{array}{c}F=\frac{11.1\text{J}/\text{K}}{13.4\text{J}/\text{K}}\\ =0.828\end{array}$

Here, $F$ is ratio of entropy in part (c) and (a).

Conclusion:

Thus the change in entropy in part (c) is less than that of part (a) by a factor of $0.828$.