Chapter 22, Problem 44P

To determine

The total entropy change of the horseshoe plus-water system.

Answer to Problem 44P

The total entropy change of the horseshoe plus-water system is $717\text{J}/\text{K}$.

Explanation of Solution

In this problem, horseshoe at $900°\text{C}$ is dropped into water at $10.0°\text{C}$.

Since there is no energy lost by heat to the surroundings, heat lost by horseshoe is equal to heat gain by water.

Write the expression for the conservation of energy.

  $Q_{\text{cold}}=-Q_{\text{hot}}$                                                                                                              (I)

Here, $Q_{\text{cold}}$ is the energy gained as heat by the cold water and $Q_{\text{hot}}$ is the energy lost by heat for hot iron.

The negative sign indicates that energy is lost by heat for hot horseshoe.

Write the expression for $Q_{\text{cold}}$.

  $Q_{\text{cold}}=m_w{c}_w\left(T_f-T_{i,w}\right)$                                                                                             (II)

Here, $m_w$ is the mass of the water, $c_w$ is the specific heat capacity of water, $T_f$ is the final temperature of the horse shoe-water system and $T_{i,w}$ is the initial temperature of the water.

Write the expression for $Q_{\text{hot}}$.

  $Q_{\text{hot}}=m_{\text{iron}}{c}_{\text{iron}}\left(T_f-T_{i,\text{iron}}\right)$                                                                                       (III)

Here, $m_{\text{iron}}$ is the mass of the iron horseshoe, $c_{\text{iron}}$ is the specific heat capacity of iron, $T_{i,\text{iron}}$ is the initial temperature of the iron horseshoe.

Use equations (II) and (III) in equation (I) to get $T_f$.

  $\begin{array}{c}{m}_w{c}_w\left(T_f-T_{i,w}\right)=-m_{\text{iron}}{c}_{\text{iron}}\left(T_f-T_{i,\text{iron}}\right)\\ T_f=\frac{m_w{c}_w{T}_{i,w}+m_{\text{iron}}{c}_{\text{iron}}{T}_{i,\text{iron}}}{m_w{c}_w+m_{\text{iron}}{c}_{\text{iron}}}\end{array}$                                                              (IV)

Write the expression for the change entropy of water.

  $\Delta S_1={\displaystyle \underset{T_{i,w}}{\overset{T_f}{\int }}\frac{m_w{c}_{w}dT}{T}}$                                                                                                     (V)

Here, $\Delta S_1$ is the entropy change of water, $T$ is the temperature and $dT$ is the differential temperature.

Write the expression for the change entropy of horseshoe.

  $\Delta S_2={\displaystyle \underset{T_{i,\text{iron}}}{\overset{T_f}{\int }}\frac{m_{\text{iron}}{c}_{\text{iron}}dT}{T}}$                                                                                              (VI)

Here, $\Delta S_2$ is the entropy change of horseshoe, $T$ is the temperature and $dT$ is the differential temperature.

Write the expression for the total entropy change of the system.

  $\Delta S=\Delta S_1+\Delta S_2$                                                                                                      (VII)

Here, $\Delta S$ is the entropy change of the system.

Put equations (V) and (VI) in equation (VII) to get $\Delta S$.

  $\Delta S={\displaystyle \underset{T_{i,w}}{\overset{T_f}{\int }}\frac{m_w{c}_{w}dT}{T}}+{\displaystyle \underset{T_{i,\text{iron}}}{\overset{T_f}{\int }}\frac{m_{\text{iron}}{c}_{\text{iron}}dT}{T}}$

Integrate above equation to get $\Delta S$.

  $\begin{array}{c}\Delta S=m_w{c}_w\ln {\left(T\right)}_{T_{i,w}}^{T_f}+m_{\text{iron}}{c}_{\text{iron}}\ln {\left(T\right)}_{T_{i,\text{iron}}}^{T_f}\\ =m_w{c}_w\ln \left(\frac{T_f}{T_{i,w}}\right)+m_{\text{iron}}{c}_{\text{iron}}\ln \left(\frac{T_f}{T_{i,\text{iron}}}\right)\end{array}$                                                             (VIII)

Write the expression to convert temperature in degree Celsius into Kelvin scale.

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$                                                                                             (IX)

Here, $T\left(\text{K}\right)$ is the temperature in Kelvin scale and $T\left(°\text{C}\right)$ is the temperature in degree Celsius.

Conclusion:

Mass of iron is $1.00\text{kg}$, mass of water is $4.00\text{kg}$, initial temperature of iron is $900°\text{C}$ and initial temperature of water is $10.0°\text{C}$.

Substitute $900.0°\text{C}$ for $T\left(°\text{C}\right)$ in equation (IX) to get $T\left(\text{K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=900.0°\text{C}+273\\ =1173\text{K}\end{array}$

Substitute $10.0°\text{C}$ for $T\left(°\text{C}\right)$ in equation (IX) to get $T\left(\text{K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=10.0°\text{C}+273\\ =283\text{K}\end{array}$

Substitute $1.00\text{kg}$ $m_{\text{iron}}$, $4.00\text{kg}$ for $m_w$, $448\text{J}/\text{kg}\cdot °\text{C}$ for $c_{\text{iron}}$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c_w$, $10.0°\text{C}$ for $T_{i,w}$ and $900°\text{C}$ for $T_{i,\text{iron}}$ in equation (IV) to get $T_f$.

  $\begin{array}{c}{T}_f=\frac{\left(4.00\text{kg}\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)\left(20.0°\text{C}\right)+\left(1.00\text{kg}\right)\left(448\text{J}/\text{kg}\cdot °\text{C}\right)\left(900°\text{C}\right)}{\left(4.00\text{kg}\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)+\left(1.00\text{kg}\right)\left(448\text{J}/\text{kg}\cdot °\text{C}\right)}\\ =33.2°\text{C}\end{array}$

Substitute $33.2°\text{C}$ for $T\left(°\text{C}\right)$ in equation (IX) to get $T\left(\text{K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=33.2°\text{C}+273\\ =306.3\text{K}\end{array}$

Substitute $1.00\text{kg}$ $m_{\text{iron}}$, $4.00\text{kg}$ for $m_w$, $448\text{J}/\text{kg}\cdot °\text{C}$ for $c_{\text{iron}}$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c_w$, $283\text{K}$ for $T_{i,w}$, $1173\text{K}$ for $T_{i,\text{iron}}$ and $306.3\text{K}$ for $T_f$ in equation (VIII) to get $\Delta S$.

  $\begin{array}{c}\Delta S=\left(4.00\text{kg}\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)\ln \left(\frac{306.3\text{K}}{283\text{K}}\right)+\left(1.00\text{kg}\right)\left(448\text{J}/\text{kg}\cdot °\text{C}\right)\ln \left(\frac{306.3\text{K}}{1173\text{K}}\right)\\ =\left(4.00\text{kg}\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)\left(0.0787\right)+\left(1.00\text{kg}\right)\left(448\text{J}/\text{kg}\cdot °\text{C}\right)\left(-1.34\right)\\ =717\text{J}/\text{K}\end{array}$

Therefore, the total entropy change of the horseshoe plus-water system is $717\text{J}/\text{K}$.