Chapter 4, Problem 12PEA

To determine

The total heat that was removed from water.

Answer to Problem 12PEA

The total heat removed from water is $40.2\text{kcal}$.

Explanation of Solution

Write the expression to calculate the change in temperature.

    $\Delta T=T_1-T_2$                                                    (I)

Here, $\Delta T$ is the change in temperature, $T_1$ is initial temperature and $T_2$ is the final temperature.

Write the expression to calculate the heat released.

    $Q_1=mc\Delta T$                                                   (II)

Here, $Q_1$ is the heat released from the water, $m$ is the mass, $c$ is the specific heat and $\Delta T$ is the change in temperature.

Write the expression to calculate the heat released for the phase change.

    $Q_2=m{L}_{\text{f}}$                                                    (III)

Here, $Q_2$ is the heat released for the phase change, $m$ is the mass and $L_{\text{f}}$ is the latent heat of fusion.

Write the expression to calculate the total heat removed from water.

    $Q_{\text{total}}=Q_1+Q_2+Q_3$                                     (IV)

Here, $Q_{\text{total}}$ is the total heat removed from water and $Q_3$ is the heat released from the ice when it cools from $0°\text{C}$ to $-5.00°\text{C}$.

Conclusion:

Case 1: The water cools from $18.0°\text{C}$ to $0°\text{C}$.

Substitute $18.0°\text{C}$ for $T_1$ and $0°\text{C}$ for $T_2$ in equation (I) to find the change in temperature.

    $\begin{array}{c}\Delta T=18.0°\text{C}-0°\text{C}\\ =18.0°\text{C}\end{array}$

Substitute $400.0\text{g}$ for $m$, $1.00\text{cal}/\text{g}°\text{C}$ for $c$ and $18.0°\text{C}$ for $\Delta T$ in equation (II) to find the heat released from the water.

    $\begin{array}{c}{Q}_1=\left(400.0\text{g}\right)\left(1.00\text{cal}/\text{g}°\text{C}\right)\left(18.0°\text{C}\right)\\ =7,200\text{cal}\left(\frac{1\text{kcal}}{1000\text{cal}}\right)\\ =7.2\text{kcal}\end{array}$

Case 2: The phase changes from water at $0°\text{C}$ to ice at $0°\text{C}$.

Substitute $400.0\text{g}$ for $m$ and $80.0\text{cal}/\text{g}$ for $L_{\text{f}}$ in equation (III) to find the heat released for the phase change.

    $\begin{array}{c}{Q}_2=\left(400.0\text{g}\right)\left(80.0\text{cal}/\text{g}\right)\\ =32,000\text{cal}\left(\frac{1\text{kcal}}{1000\text{cal}}\right)\\ =32.0\text{kcal}\end{array}$

Case 3: The ice cool from $0°\text{C}$ to $-5.00°\text{C}$.

Substitute $0°\text{C}$ for $T_1$ and $-5.00°\text{C}$ for $T_2$ in equation (I) to find the change in temperature.

    $\begin{array}{c}\Delta T=0°\text{C}-\left(-5.00°\text{C}\right)\\ =5.00°\text{C}\end{array}$

Substitute $400.0\text{g}$ for $m$, $0.50\text{cal}/\text{g}°\text{C}$ for $c$ and $5.00°\text{C}$ for $\Delta T$ in equation (II) to find the energy released from the ice.

    $\begin{array}{c}{Q}_3=\left(400.0\text{g}\right)\left(0.50\text{cal}/\text{g}°\text{C}\right)\left(5.00°\text{C}\right)\\ =1,000\text{cal}\left(\frac{1\text{kcal}}{1000\text{cal}}\right)\\ =1.0\text{kcal}\end{array}$

Substitute $7.2\text{kcal}$ for $Q_1$, $32.0\text{kcal}$ for $Q_2$ and $1.0\text{kcal}$ for $Q_3$ in equation (IV) to find the total heat removed from water.

    $\begin{array}{c}{Q}_{\text{total}}=7.2\text{kcal}+32.0\text{kcal}+1.0\text{kcal}\\ =40.2\text{kcal}\end{array}$

Therefore, the total heat removed from water is $40.2\text{kcal}$.