Chapter 14, Problem 49P

To determine

The final temperature of the system.

Answer to Problem 49P

The final temperature of the system is $\underset{\_}{T=32°\text{C}}$.

Explanation of Solution

Write the expression for heat required to melt ice.

$Q=m_{\text{ice}}{L}_{\text{f}}+m_{\text{ice}}{c}_{\text{ice}}\Delta T$ (I)

Here, $Q$ is the heat, $m_{\text{ice}}$ is the mass, $c_{\text{ice}}$ is the specific heat capacity of ice, $\Delta T$ is the temperature, and $L_{\text{f}}$ is the latent heat of fusion.

Write the expression for heat absorbed from water.

$\left|Q\right|=m_{\text{w}}{c}_{\text{w}}\left|\Delta T\right|$ (II)

Here, $m_{\text{w}}$ is the mass of water, and $c_{\text{w}}$ is the specific heat capacity of water

Write the expression for the total heat flow in the system.

$\begin{array}{c}{Q}_{\text{w}}+Q_{\text{ice}}=0\\ m_{\text{w}}{c}_{\text{w}}\Delta T_{\text{w}}+m_{\text{ice}}{L}_{\text{f}}+m_{\text{ice}}{c}_{\text{ice}}\Delta T_1+m_{\text{ice}}{c}_{\text{w}}\Delta T_2=0\end{array}$ (III)

Substitute $T-T_{\text{w}}$ for $\Delta T_{\text{w}}$, $\left(273.15\text{K}-T_{\text{ice}}\right)$ for $\Delta T_1$, and $\left(T-273.15\text{K}\right)$ for $\Delta T_2$ in equation (III) and rearrange to obtain an expression for $T$.

$\begin{array}{c}{m}_{\text{ice}}{c}_{\text{ice}}\left(273.15\text{K}-T_{\text{ice}}\right)+m_{\text{ice}}{c}_{\text{w}}\left(T-273.15\text{K}\right)=-m_{\text{w}}{c}_{\text{w}}\left(T-T_{\text{w}}\right)+m_{\text{ice}}{L}_{\text{f}}\\ \left(m_{\text{w}}{c}_{\text{w}}+m_{\text{ice}}{c}_{\text{w}}\right)T+m_{\text{ice}}{L}_{\text{f}}+m_{\text{ice}}\left(c_{\text{ice}}-c_{\text{w}}\right)\left(273.15\text{K}\right)=m_{\text{w}}{c}_{\text{w}}{T}_{\text{w}}+m_{\text{ice}}{c}_{\text{ice}}{T}_{\text{ice}}=0\\ T=\frac{m_{\text{w}}{c}_{\text{w}}{T}_{\text{w}}+m_{\text{ice}}{c}_{\text{ice}}{T}_{\text{ice}}-m_{\text{ice}}{L}_{\text{f}}-m_{\text{ice}}\left(c_{\text{ice}}-c_{\text{w}}\right)\left(273.15\text{K}\right)}{c_{\text{w}}\left(m_{\text{w}}+m_{\text{ice}}\right)}\end{array}$ (IV)

Conclusion:

Substitute, $0.075\text{kg}$ for $m_{\text{ice}}$, $333.7\text{kJ/kg}$ for $L_{\text{f}}$, $2.1\text{kJ/kg}\cdot \text{K}$ for $c_{\text{ice}}$, and $10\text{K}$ $\Delta T$ in equation (I).

$\begin{array}{c}Q=\left(0.075\text{kg}\right)\left\{333.7\text{kJ/kg}+\left[2.1\text{kJ/kg}\cdot \text{K}\right]\left(10\text{K}\right)\right\}\\ =27\text{kJ}\end{array}$

Substitute, $0.50\text{kg}$ for $m_{\text{w}}$, $4.186\text{kJ/kg}\cdot \text{K}$ for $c_{\text{w}}$, $50\text{K}$ for $\Delta T$ in equation (II).

$\begin{array}{c}\left|Q\right|=0.50\text{kg}\times 4.186\text{kJ/kg}\cdot \text{K}\times 50\text{K}\\ \text{=105}\text{kJ}\end{array}$

Since, the $105\text{kJ>27}\text{kJ}$ ice will melt completely.

Substitute, $0.50\text{kg}$ for $m_{\text{w}}$, $4.186\text{kJ/kg}\cdot \text{K}$ for $c_{\text{w}}$, $\left(50.0+273.15\right)\text{K}$ for $T_{\text{w}}$, $0.075\text{kg}$ for $m_{\text{ice}}$, $\left(-10+273.15\right)\text{K}$ for $T_{\text{ice}}$, $333.7\text{kJ/kg}$ for $L_{\text{f}}$, $2.1\text{kJ/kg}\cdot \text{K}$ for $c_{\text{ice}}$ in equation (IV)

$\begin{array}{c}T=\frac{\left(0.50\text{kg}\right)\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(\left(50.0+273.15\right)\text{K}\right)+\left(0.075\text{kg}\right)\left\{\left[2.1\text{kJ/kg}\cdot \text{K}\right]\left(-10+273.15\right)\text{K}-333.7\text{kJ/kg}\right\}-\left(0.075\text{kg}\right)\left[2.1\text{kJ/kg}\cdot \text{K}-4.186\text{kJ/kg}\cdot \text{K}\right]\left(273.15\text{K}\right)}{\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(0.50\text{kg}+0.075\text{kg}\right)}\\ =305.2\text{K}-\text{273}\text{.15}\text{K}\\ \text{=}\text{32}°\text{C}\end{array}$

Therefore, the final temperature of the system is $\underset{\_}{T=32°\text{C}}$.