Chapter 4, Problem 4I.5E

Interpretation Introduction

Interpretation:

The value of $\Delta \text{S}$ and $\Delta {\text{S}}_{total}$ for the process have to be determined.

Concept Introduction:

The heat capacity is the ratio of the energy supplied in the form of heat to the rise in the temperature of the system.  The mathematical relation for the heat capacity is shown below.

  $C=\frac{q}{\Delta T}$

The degree of randomness in a system is the predicted according to its respective entropy.  Higher the entropy greater will be the disorder in the system.  The mathematical expression for the calculation of entropy of the system is shown below.

  $\Delta \text{S}=\frac{{\text{q}}_{rev}}{\text{T}}$

Explanation of Solution

The mass of water at $20°\text{C}$ is $50\text{g}$.  However, the mass of water at $50°\text{C}$ is $\text{65}\text{g}$.  The specific heat capacity of water is $4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}$.

The unit conversion of temperature of two different samples of water from $°\text{C}$ to $\text{K}$ is shown below.

    $\begin{array}{c}{T}_{\text{1}}=20+273\text{K}\\ =293\text{K}\\ T_{\text{2}}=50+273\text{K}\\ =323\text{K}\end{array}$

As per the given data in the question, there is no energy lost to the surroundings.  Therefore, the heat loss by the water at $20°\text{C}$ is equal to the heat gained by the water at $50°\text{C}$.  The mathematical expression representing this is shown below.

    $m_{\text{1}}{C}_{\text{s}}\left(T_1-T\right)=m_{\text{2}}{C}_{\text{s}}\left(T-T_{\text{2}}\right)$        (1)

Where,

• $m_{\text{1}}$ is the mass of first sample.
• $m_{\text{2}}$ is the mass of second sample.
• $C_{\text{s}}$ is the specific heat capacity of water.
• $T_{\text{1}}$ is the initial temperature of first sample.
• $T_{\text{2}}$ is the initial temperature of second sample.
• $T$ is the final temperature of the system.

The value of $m_{\text{1}}$ is $50\text{g}$.

The value of $m_{\text{2}}$ is $65\text{g}$.

The value of $C_{\text{s}}$ is $4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}$.

The value of $T_{\text{1}}$ is $293\text{K}$.

The value of $T_{\text{2}}$ is $323\text{K}$.

Substitute the value of $m_{\text{1}}$, $m_{{\text{H}}_{\text{2}}\text{O}}$, $C_{\text{s}}$, $T_{\text{1}}$ and $T_{\text{2}}$ in equation (1).

    $\begin{array}{c}\left(50\text{g}\right)\times \left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}\right)\times \left(293\text{K}-T\right)=\left(65\text{g}\right)\times \left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}\right)\times \left(T-323\text{K}\right)\\ \left(50\text{g}\right)\times \left(293\text{K}-T\right)=\left(65\text{g}\right)\times \left(T-323\text{K}\right)\\ 14650\text{K}-50T=65T-20995\text{K}\\ 65T+50T=14650\text{K}+20995\text{K}\end{array}$

On further calculation the final temperature of the system is calculated as shown below.

    $\begin{array}{c}115T=35645\text{K}\\ T=\frac{35645}{115}\text{K}\\ =309.95\text{K}\end{array}$

Therefore, the final temperature of the system is $309.95\text{K}$.

The relation to calculate the change in the entropy of the water at $20°\text{C}$ is shown below.

    $\Delta \text{S}=\text{m}{C}_s\ln \left(\frac{{\text{T}}_f}{{\text{T}}_i}\right)$        (2)

Where,

• $\Delta \text{S}$ is change in entropy.
• $m$ is the mass of sample.
• $C_{\text{s}}$ is the specific heat capacity.
• $T_{\text{i}}$ is the initial temperature of the system.
• $T_{\text{f}}$ is the final temperature of the system.

The value of $m$ is $50\text{g}$.

The value of $C_{\text{s}}$ is $4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}$.

The value of $T_{\text{i}}$ is $293\text{K}$.

The value of $T_{\text{f}}$ is $309.95\text{K}$.

Substitute the value of $m$, $C_{\text{s}}$, $T_{\text{i}}$ and $T_{\text{f}}$ in equation (2).

    $\begin{array}{c}\Delta \text{S}=\left(50\text{g}\right)\times \left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}\right)\times \ln \left(\frac{309.95\text{K}}{293\text{K}}\right)\\ =\left(209.2\text{J}\cdot {\text{K}}^{-1}\right)\times \ln \left(1.0578\right)\\ =\left(209.2\text{J}\cdot {\text{K}}^{-1}\right)\times \left(0.05619\right)\\ =+11.75\text{J}\cdot {\text{K}}^{-1}\end{array}$

Therefore, the change in the entropy of the water at $20°\text{C}$ is $+11.75\text{J}\cdot {\text{K}}^{-1}$.

The relation to calculate the change in the entropy of the water at $50°\text{C}$ is same as equation (2).

The value of $m$ is $65\text{g}$.

The value of $C_{\text{s}}$ is $4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}$.

The value of $T_{\text{i}}$ is $323\text{K}$.

The value of $T_{\text{f}}$ is $309.95\text{K}$.

Substitute the value of $m$, $C_{\text{s}}$, $T_{\text{i}}$ and $T_{\text{f}}$ in equation (2).

    $\begin{array}{c}\Delta \text{S}=\left(65\text{g}\right)\times \left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot {\text{K}}^{-1}\right)\times \ln \left(\frac{309.95\text{K}}{323\text{K}}\right)\\ =\left(271.96\text{J}\cdot {\text{K}}^{-1}\right)\times \ln \left(0.9595\right)\\ =\left(271.96\text{J}\cdot {\text{K}}^{-1}\right)\times \left(-0.0413\right)\\ =-11.23\text{J}\cdot {\text{K}}^{-1}\end{array}$

Therefore, the change in the entropy of the water at $50°\text{C}$ is $-11.23\text{J}\cdot {\text{K}}^{-1}$.

The relation to calculate the total change in the entropy is shown below.

    $\Delta {\text{S}}_{total}=\Delta \text{S}\left(20°\text{C}\right)+\Delta \text{S}\left(50°\text{C}\right)$        (3)

Where,

• $\Delta {\text{S}}_{total}$ is the total change in the entropy.
• $\Delta \text{S}\left(20°\text{C}\right)$ is the change in the entropy at $20°\text{C}$.
• $\Delta \text{S}\left(50°\text{C}\right)$ is the change in the entropy at $50°\text{C}$.

The value of $\Delta \text{S}\left(20°\text{C}\right)$ is $+11.75\text{J}\cdot {\text{K}}^{-1}$.

The value of $\Delta \text{S}\left(50°\text{C}\right)$ is $-11.23\text{J}\cdot {\text{K}}^{-1}$.

Substitute the value of $\Delta \text{S}\left(20°\text{C}\right)$ and $\Delta \text{S}\left(50°\text{C}\right)$ in equation (3).

  $\begin{array}{c}\Delta {\text{S}}_{total}=+11.75\text{J}\cdot {\text{K}}^{-1}-11.23\text{J}\cdot {\text{K}}^{-1}\\ =0.52\text{J}\cdot {\text{K}}^{-1}\end{array}$

Therefore, the total change in the entropy is $0.52\text{J}\cdot {\text{K}}^{-1}$.

The mixture of two different samples of water is in the insulated vessel.  The thermally insulated vessel is considered as the isolated system as there is neither exchange of energy nor matter takes place from it.  Therefore, the value of $\Delta {\text{S}}_{surr}$ is zero.

The relation to calculate $\Delta \text{S}$ is shown below.

    $\Delta \text{S}=\Delta {\text{S}}_{total}-\Delta {\text{S}}_{surr}$        (4)

Where,

• $\Delta {\text{S}}_{total}$ is the total change in the entropy.
• $\Delta \text{S}$ is the change in the entropy of system.
• $\Delta {\text{S}}_{surr}$ is the change in the entropy of surrounding.

The value of $\Delta {\text{S}}_{total}$ is $0.52\text{J}\cdot {\text{K}}^{-1}$.

The value of $\Delta {\text{S}}_{surr}$ is $0.0\text{J}\cdot {\text{K}}^{-1}$.

Substitute the value of $\Delta {\text{S}}_{total}$ and $\Delta {\text{S}}_{surr}$ in equation (4).

    $\begin{array}{c}\Delta \text{S}=0.52\text{J}\cdot {\text{K}}^{-1}-0.0\text{J}\cdot {\text{K}}^{-1}\\ =0.52\text{J}\cdot {\text{K}}^{-1}\end{array}$

Thus, the value of both $\Delta \text{S}$ and $\Delta {\text{S}}_{total}$ for the process is $\underset{\_}{0.52J\cdot K^{-1}}$.