Chapter 3, Problem 3.35E

A sample of $1.00\text{mol}$ of silver at $150°\text{C}$ is placed in contact with $1.00\text{mol}$ of silver at $0°\text{C}$. Calculate (a) the final temperature of both silver samples; (b) The $\Delta S$ for the hot $\text{Ag}$ sample; (c) the $\Delta S$ for the cold $\text{Ag}$ sample; and (d) the total $\Delta S$ of the system. (e) Is the process spontaneous? How do you know? Assume a constant heat capacity for $\text{Ag}$ of $25.75\text{J}/\text{mol}\cdot \text{K}$.

Interpretation Introduction

(a)

Interpretation:

The final temperature of the hot and cold silver is to be calculated.

Concept introduction:

The first law of thermodynamics states that the total energy of an isolated system remains unchanged. The molar heat capacity of a substance is defined as the amount of heat required to raise the temperature of one mole of that substance by unity. The molar heat capacity of a substance is shown below.

$C=\frac{q}{n\left(T_{\text{f}}-T_{\text{i}}\right)}$

Answer to Problem 3.35E

The final temperature of the hot and cold silver is $\text{75}°\text{C}$.

Explanation of Solution

The initial temperature of the hot silver is $150°\text{C}$.

The initial temperature of hot silver in Kelvin is shown below.

$\begin{array}{rll}T& =& \left(273+150°\text{C}\right)\text{K}\\ & =& 423\text{K}\end{array}$

The initial temperature of cold silver is $0°\text{C}$.

The initial temperature of cold silver in Kelvin is shown below as,

$\begin{array}{rll}T& =& \left(273+0°\text{C}\right)\text{K}\\ & =& 273\text{K}\end{array}$

The final temperature of both cold silver and hot silver is the same and assumed to be $T_{\text{f}}$.

The number of moles of both cold and hot silver is $1.00\text{mol}$.

The molar heat capacity of silver is $25.75\text{J}/\text{mol}\cdot \text{K}$.

The exchange of heat due to temperature change is shown below.

$q=nC\left(T_{\text{f}}-T_{\text{i}}\right)$ …(1)

Where,

• $n$ represents the number of moles of the substance.

• $C$ represents the molar heat capacity of the substance.

• $T_{\text{f}}$ represents the final temperature of the substance.

• $T_{\text{i}}$ represents the initial temperature of the substance.

In terms of the first of law of thermodynamics, the energy released by hot silver is equal to the heat accepted by cold silver. The relationship between heat exchanges is shown below.

$q_{\text{c}}=-q_{\text{h}}$ …(2)

Where,

• $q_{\text{h}}$ represents the heat change of hot silver.

• $q_{\text{c}}$ represents the heat change of cold silver.

Substitute equation (1) in equation (2).

$n_{\text{h}}{C}_{\text{h}}\left(T_{\text{f}}-T_{\text{i, h}}\right)=-n_{\text{c}}{C}_{\text{c}}\left(T_{\text{f}}-T_{\text{i, c}}\right)$ …(3)

Rearrange the equation (3) for the value of $T_{\text{f}}$.

$T_{\text{f}}=\frac{n_{\text{c}}{C}_{\text{c}}{T}_{\text{i, c}}+n_{\text{h}}{C}_{\text{h}}{T}_{\text{i, h}}}{n_{\text{c}}{C}_{\text{c}}+n_{\text{h}}{C}_{\text{h}}}$ …(4)

Substitute the values of initial temperature, molar heat capacity and number of moles of hot silver and cold silver in equation (4).

$\begin{array}{rll}{T}_{\text{f}}& =& \frac{\left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)\left(273\text{K}\right)+\left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)\left(423\text{K}\right)}{\left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)+\left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)}\\ & =& \left(\frac{\left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)}{\left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)}\right)\left(\frac{273\text{K}+423\text{K}}{2}\right)\text{K}\\ & =& \text{348}\text{K}\end{array}$

The final temperature of the hot and cold silver is $\text{348}\text{K}$.

The final temperature of the hot and cold silver in degree Celsius is represented as,

$\begin{array}{rll}T& =& \left(\text{348}\text{K}-273\right)°\text{C}\\ & =& \text{75}°\text{C}\end{array}$

Therefore, the final temperature of the hot and cold silver is $\text{75}°\text{C}$.

Conclusion

The final temperature of the hot and cold silver is $\text{75}°\text{C}$.

Interpretation Introduction

(b)

Interpretation:

The entropy change of the hot $\text{Ag}$ sample is to be calculated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases. The second law of thermodynamics states that the entropy of the system either increases or remains the same.

Answer to Problem 3.35E

The entropy change of hot $\text{Ag}$ sample is $-5.03\text{J}/\text{K}$.

Explanation of Solution

The initial temperature of the hot silver is $150°\text{C}$.

The initial temperature of hot silver in Kelvin is shown below.

$\begin{array}{rll}T& =& \left(273+150°\text{C}\right)\text{K}\\ & =& 423\text{K}\end{array}$

The final temperature of the hot silver is $\text{348}\text{K}$.

The number of moles of hot silver is $1.00\text{mol}$.

The molar heat capacity of silver is $25.75\text{J}/\text{mol}\cdot \text{K}$.

The entropy change for the temperature change is shown below.

$\Delta S=nC\ln \left(\frac{T_{\text{f}}}{T_{\text{i}}}\right)$ …(5)

Where,

• $n$ represents the number of moles of the substance.

• $C$ represents the molar heat capacity of the substance.

• $T_{\text{f}}$ represents the final temperature of the substance.

• $T_{\text{i}}$ represents the initial temperature of the substance.

Substitute the values of final temperature, initial temperature, molar heat capacity and mass of hot silver in equation (5).

$\begin{array}{rll}\Delta S& =& \left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)\ln \left(\frac{\text{348}\text{K}}{423\text{K}}\right)\\ & =& -5.0256\text{J}/\text{K}\\ & \approx & -5.03\text{J}/\text{K}\end{array}$

Therefore, the entropy change of the hot $\text{Ag}$ sample is $-5.03\text{J}/\text{K}$.

Conclusion

The entropy change of hot $\text{Ag}$ sample is $-5.03\text{J}/\text{K}$.

Interpretation Introduction

(c)

Interpretation:

The entropy change of the cold silver sample is to be calculated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases. The second law of thermodynamics states that the entropy of the system either increases or remains the same.

Answer to Problem 3.35E

The entropy change of the cold $\text{Ag}$ sample is $6.25\text{J}/\text{K}$.

Explanation of Solution

The initial temperature of cold silver is $0°\text{C}$.

The initial temperature of cold silver in Kelvin is shown below.

$\begin{array}{rll}T& =& \left(273+0°\text{C}\right)\text{K}\\ & =& 273\text{K}\end{array}$

The final temperature of the cold silver is $\text{348}\text{K}$.

The number of moles of cold silver is $1.00\text{mol}$.

The molar heat capacity of silver is $25.75\text{J}/\text{mol}\cdot \text{K}$.

The entropy change for the temperature change is shown below.

$\Delta S=nC\ln \left(\frac{T_{\text{f}}}{T_{\text{i}}}\right)$ …(5)

Where,

• $n$ represents the number of moles of the substance.

• $C$ represents the molar heat capacity of the substance.

• $T_{\text{f}}$ represents the final temperature of the substance.

• $T_{\text{i}}$ represents the initial temperature of the substance.

Substitute the values of final temperature, initial temperature, molar heat capacity and mass of cold silver in equation (5).

$\begin{array}{rll}\Delta S& =& \left(1.00\text{mol}\right)\left(25.75\text{J}/\text{mol}\cdot \text{K}\right)\ln \left(\frac{\text{348}\text{K}}{273\text{K}}\right)\\ & =& 6.25\text{J}/\text{K}\end{array}$

Therefore, the entropy change of the cold $\text{Ag}$ sample is $6.25\text{J}/\text{K}$.

Conclusion

The entropy change of the cold $\text{Ag}$ sample is $6.25\text{J}/\text{K}$.

Interpretation Introduction

(d)

Interpretation:

The total entropy change in the system of silver is to be calculated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases. The second law of thermodynamics states that the entropy of the system either increases or remains the same.

Answer to Problem 3.35E

The total entropy change in the system of silver is $1.22\text{J}/\text{K}$.

Explanation of Solution

The entropy change of hot $\text{Ag}$ sample is $-5.03\text{J}/\text{K}$.

The entropy change of the cold $\text{Ag}$ sample is $6.25\text{J}/\text{K}$.

The total entropy change of the given mixture is shown below.

$\Delta S_{\text{t}}=\Delta S_{\text{c}}+\Delta S_{\text{h}}$

Where,

• $\Delta S_{\text{h}}$ represents the entropy change of hot silver.

• $\Delta S_{\text{c}}$ represents the entropy change of cold silver.

Substitute the values of $\Delta S_{\text{h}}$ and $\Delta S_{\text{c}}$ in the above equation.

$\begin{array}{rll}\Delta S_{\text{t}}& =& 6.25\text{J}/\text{K}+\left(-5.03\text{J}/\text{K}\right)\\ & =& 1.22\text{J}/\text{K}\end{array}$

Therefore, the total entropy change in the system of silver is $1.22\text{J}/\text{K}$.

Conclusion

The total entropy change in the system of silver is $1.22\text{J}/\text{K}$.

Interpretation Introduction

(e)

Interpretation:

Whether the process of heat exchange is spontaneous or not is to be stated. The reason for the corresponding answer is to be stated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less order arrangement, then the entropy of the system increases. The second law of thermodynamics state that the entropy of the system either increases or remain the same.

Answer to Problem 3.35E

The process is spontaneous because the entropy change of the system is positive. According to the second law of thermodynamics, the entropy change of system must be positive for a spontaneous process.

Explanation of Solution

The total entropy change in the system of silver is $1.22\text{J}/\text{K}$.

This value implies that the randomness in the system is increasing with time. The second law of thermodynamics states that the entropy of the system either increases or remains the same. The entropy change is positive that means the entropy of the system is increasing. Therefore, the process is spontaneous.

Conclusion

The process is spontaneous because the entropy change of the system is positive. According to the second law of thermodynamics, the entropy change of system must be positive for a spontaneous process.