Chapter 18, Problem 30P

To determine

The change in entropy when a $27.9\text{g}$ of ice cube at $-12°\text{C}$ is transformed into steam at $115°\text{C}$.

Answer to Problem 30P

The change in entropy when a $27.9\text{g}$ of ice cube at $-12°\text{C}$ is transformed into steam at $115°\text{C}$ is $\text{244}\text{.5}\text{J/K}$.

Explanation of Solution

Given Info: The mass of the ice cube is $27.9\text{g}$, the value of cold temperature is $-12°\text{C}$ and the value of value of hot temperature is $115°\text{C}$.

Formula to find change in entropy to transform ice at $-12°\text{C}$  to ice at $0°\text{C}$  is,

$\Delta S_1=mC\text{ln}\left(\frac{T_{\text{h}}}{T_{\text{c}}}\right)$

Here,

$\Delta S_1$ is the entropy change to transform ice at $-12°\text{C}$  to ice at $0°\text{C}$.

$m$ is the mass of ice.

$T_{\text{h}}$ is the hot temperature.

$T_{\text{c}}$ is the cold temperature.

$C$ is the specific heat of ice

Substitute $27.9\text{g}$ for $m$, $2.108\text{J}/\text{g}\cdot \text{K}$ for $C$, $0\text{°C}$ for $T_{\text{h}}$ and $-12°\text{C}$ for $T_{\text{c}}$ to find the change in entropy.

$\begin{array}{c}\Delta S_1=\left(27.9\text{g}\times 2.108\text{J}/\text{g}\cdot \text{K}\times \ln \left(\frac{\left(273+0\right)\text{K}}{\left(273-12\right)\text{K}}\right)\right)\\ =\left(27.9\times 2.108\times \ln \left(\frac{273}{261}\right)\right)\\ =2.64\text{J}/\text{K}\end{array}$

Thus, the change in the entropy to transform ice at $-12°\text{C}$ to ice at $0°\text{C}$ is $2.64\text{J}/\text{K}$.

Formula to find the change in entropy to transform ice at $0°\text{C}$  to water at $0°\text{C}$ is,

$\Delta S_2=m\frac{L}{T_1}$

Here,

$\Delta S_2$ is the entropy change to transform to transform ice at $0°\text{C}$  to water at $0°\text{C}$.

$m$ is the mass of water.

$T_1$ is the transition temperature from ice to water.

$L$ is the latent heat of fusion

Substitute $27.9\text{g}$ for $m$, $334\text{J}/\text{g}$ for $L$ and $0\text{°C}$ for $T_1$ to find the change in entropy.

$\begin{array}{c}\Delta S_2=\left(27.9\text{g}\times \frac{334\text{J}/\text{g}}{\left(273+0\right)\text{K}}\right)\\ =\left(27.9\times \frac{344}{273}\right)\text{J}/\text{K}\\ =34.1\text{J}/\text{K}\end{array}$

Thus, the change in the entropy to transform ice at $0°\text{C}$  to water at $0°\text{C}$ is $34.1\text{J}/\text{K}$.

Formula to find change in entropy to transform water at $0°\text{C}$ to water at $100°\text{C}$ is,

$\Delta S_3=mC\text{ln}\left(\frac{T_h}{T_c}\right)$

Here,

$\Delta S_3$ is the entropy change to transform water at $0°\text{C}$ to water at $100°\text{C}$.

$m$ is the mass of water.

$C$ is the specific heat of water

Substitute $27.9\text{g}$ for $m$, $4.19\text{J}/\text{g}\cdot \text{K}$ for $C$, $100\text{°C}$ for $T_{\text{h}}$ and $0°\text{C}$ for $T_{\text{c}}$ to find the change in entropy.

$\begin{array}{c}\Delta S_3=\left(27.9\text{g}\times 4.19\text{J}/\text{gK}\times \ln \left(\frac{273+100\text{K}}{273+0\text{K}}\right)\right)\\ =\left(27.9\times 4.19\times \ln \left(\frac{373}{273}\right)\right)\\ =36.46\text{J/K}\end{array}$

Thus, the change in the entropy to transform water at $0°\text{C}$ to water at $100°\text{C}$ is $36.46\text{J}/\text{K}$.

Formula of find change in entropy to transform water at $100°\text{C}$ to steam at $100°\text{C}$ is,

$\Delta S_4=m\frac{L}{T_2}$

Here,

$\Delta S_4$ is the entropy change to transform water at $100°\text{C}$ to steam at $100°\text{C}$.

$m$ is the mass of water.

$T_2$ is the transition temperature from water to steam.

$L$ is the latent heat of vaporization.

Substitute $27.9\text{g}$ for $m$, $2256\text{J}/\text{g}$ for $L$ and $100\text{°C}$ for $T_2$ to find the change in entropy.

$\begin{array}{c}\Delta S_4=\left(27.9\text{g}\times \frac{2256\text{J}/\text{g}}{273+100\text{K}}\right)\\ =\left(27.9\times \frac{2256}{373}\right)\text{J}/\text{K}\\ =168.7\text{J}/\text{K}\end{array}$

Thus, the change in the entropy to transform water at $100°\text{C}$ to steam at $100°\text{C}$ is $168.7\text{J}/\text{K}$.

Formula to find change in entropy to transform steam at $100°\text{C}$ to steam at $115°\text{C}$ is,

$\Delta S_5=mC\text{ln}\left(\frac{T_h}{T_c}\right)$

Here,

$\Delta S_5$ is the entropy change to transform steam at $100°\text{C}$ to steam at $115°\text{C}$.

$m$ is the mass of steam.

$T_h$ is the hot temperature.

$T_c$ is the cold temperature.

$C$ is the specific heat of steam

Substitute $27.9\text{g}$ for $m$, $1.99\text{J}/\text{g}\cdot \text{K}$ for $C$, $115\text{°C}$ for $T_{\text{h}}$ and $100°\text{C}$ for $T_{\text{c}}$ to find the change in entropy.

$\begin{array}{c}\Delta S_5=\left(27.9\text{g}\times 1.99\text{J}/\text{g}\cdot \text{K}\times \ln \left(\frac{273+115\text{K}}{273+100\text{K}}\right)\right)\\ =\left(27.9\times 1.99\times \ln \left(\frac{388}{373}\right)\right)\text{J}/\text{K}\\ =2.19\text{J}/\text{K}\end{array}$

Thus, the change in the entropy to transform steam at $100°\text{C}$ to steam at $115°\text{C}$ is $2.19\text{J}/\text{K}$.

Formula to find the total change in entropy to transform ice at $-12°\text{C}$ to steam at $115°\text{C}$ is,

$\Delta S=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Substitute $2.64\text{J}/\text{K}$ for $\Delta S_{\text{1}}$, $34.1\text{J}/\text{K}$ for $\Delta S_2$, $36.46\text{J}/\text{K}$ for $\Delta S_3$, $168.7\text{J}/\text{K}$ for $\Delta S_4$ and $2.19\text{J}/\text{K}$ for $\Delta S_5$ in above expression.

$\begin{array}{c}\Delta S=2.64\text{J}/\text{K}+34.1\text{J}/\text{K}+36.46\text{J}/\text{K}+168.7\text{J}/\text{K}+2.19\text{J}/\text{K}\\ =244.1\text{J}/\text{K}\end{array}$

Thus, the total change in the entropy is $\text{244}\text{.1}\text{J/K}$.

Conclusion:

Therefore, the total change in the entropy is $\text{244}\text{.1}\text{J/K}$.