Chapter 12, Problem 80A

(a)

To determine

The change in total entropy of the reservoir.

Answer to Problem 80A

The change in total entropy of the reservoir is

Explanation of Solution

Given:

The heat energy removed from the hot reservoir is

The temperature of the hot reservoir is

The thermal energy transferred to the colder reservoir is $Q_{\text{C}}=40.0\text{J}$ .

The temperature of the colder reservoir is $T_{\text{C}}=325\text{K}$ .

The entropy is also transferred from the one reservoir to the other.

Formula used:

The expression for the entropy change is,

$\Delta S=\frac{Q}{T}$

Here, is the heat added to the object and is the temperature of the object in kelvins.

Calculation:

The net increase in the entropy of the reservoirs is,

$\begin{array}{l}\Delta S_{\text{T}}=\Delta S_{\text{C}}-\Delta S_{\text{H}}\\ \Delta S_{\text{T}}=\frac{Q_{\text{C}}}{T_{\text{C}}}-\frac{Q_{\text{H}}}{T_{\text{H}}}\\ \Delta S_{\text{T}}=\frac{40.0\text{J}}{325\text{K}}-\frac{50.0\text{J}}{545\text{K}}\\ \Delta S_{\text{T}}=0.031\text{J/K}\end{array}$

Conclusion:

Thus, the change in total entropy of the reservoir is $0.031\text{J/K}$ .

(b)

To determine

The total entropy change of the reservoir for the given temperature.

Answer to Problem 80A

The total entropy change of the reservoir is $0.103\text{J/K}$ .

Explanation of Solution

Given:

The heat energy removed from the hot reservoir is

The temperature of the hot reservoir is

The thermal energy transferred to the colder reservoir is $Q_{\text{C}}=40.0\text{J}$ .

The temperature of the colder reservoir is $T_{\text{C}}=205\text{K}$ .

The entropy is also transferred from the one reservoir to the other.

Formula used:

The expression for the entropy change is

$\Delta S=\frac{Q}{T}$

Here, is the heat added to the object and is the temperature of the object in kelvins.

Calculation:

The net increase in the entropy of the reservoirs is,

$\begin{array}{l}\Delta S_{\text{T}}=\Delta S_{\text{C}}-\Delta S_{\text{H}}\\ \Delta S_{\text{T}}=\frac{Q_{\text{C}}}{T_{\text{C}}}-\frac{Q_{\text{H}}}{T_{\text{H}}}\\ \Delta S_{\text{T}}=\frac{40.0\text{J}}{205\text{K}}-\frac{50.0\text{J}}{545\text{K}}\\ \Delta S_{\text{T}}=0.103\text{J/K}\end{array}$

Conclusion:

Thus, the total entropy change of the reservoir if the temperature of the colder reservoir is is $0.103\text{J/K}$ .