Chapter 11, Problem 35P

To determine

The final temperature of the system consisting of the ice, water and calorimeter.

Answer to Problem 35P

The final temperature is $\text{16}°\text{C}$.

Explanation of Solution

Given Info: Mass of ice block is 40 g, initial temperature of ice block is $-78°\text{C}$,  mass of water is 560 g, mass of copper calorimeter is 80 g, initial temperature of water and calorimeter is $25°\text{C}$, and Specific heat of ice is $2090\text{J}/\text{kg}\cdot \text{°C}$

There are two possibilities; the ice can melt completely or partially. This can be identified only with the final temperature of the mixture. If the final temperature is greater than $0°\text{C}$ then all the ice melts else only partially melts. To find the final temperature let us assume that the ice melts completely.

Heat energy lost be ice is equal to heat energy gained by water and calorimeter.

Formula to calculate the heat required to raise the temperature of ice $0°\text{C}$ is,

$Q_{\text{ice}}=m_{\text{ice}}{c}_{\text{ice}}\left(T_f{,}_{\text{ice}}-T_i{,}_{\text{ice}}\right)$ (1)

• $Q_{\text{ice}}$ is the heat required to raise the temperature of ice to $0°\text{C}$,
• $m_{\text{ice}}$ is the mass of the ice,
• $T_i{,}_{\text{ice}}$ is the initial temperature of ice,
• $T_f{,}_{\text{ice}}$ is the final temperature of ice,

Formula to calculate the heat required to melt the ice to cold water is,

$Q_{\text{melt}}=m_{\text{ice}}{L}_f$ (2)

• $Q_{\text{melt}}$ is the energy required to melt the ice to cold water,
• $m_{\text{ice}}$ is the mass of the ice,
• $L_f$ is the latent heat of fusion of ice,

Formula to calculate the heat required to raise the temperature of cold water.

$Q_{\text{ice-water}}=m_{\text{ice}}{c}_{\text{water}}\left(T_f-T_i{,}_{\text{ice-water}}\right)$ (3)

• $Q_{\text{ice-water}}$ is the heat required to raise the temperature of ice water.
• $c_{\text{water}}$ is the specific heat of water,
• $T_i{,}_{\text{ice-water}}$ is the initial temperature of cold water from ice,
• $T_f$ is the final temperature of the mixture.

Formula to calculate the heat lost by the water is,

$Q_{\text{water}}=m_{\text{water}}{c}_{\text{water}}\left(T_f-T_i{,}_{\text{water}}\right)$ (4)

• $Q_{\text{water}}$ is the heat lost be the water,
• $c_{\text{water}}$ is the specific heat of water,
• $T_i{,}_{\text{water}}$ is the initial temperature of cold water from ice,

Formula to calculate the heat lost by the calorimeter is,

$Q_{\text{cal}}=m_{\text{cal}}{c}_{\text{cu}}\left(T_f-T_i{,}_{\text{cal}}\right)$ (5)

• $Q_{\text{cal}}$ is the heat lost by the calorimeter,
• $m_{\text{cal}}$ is the mass of the calorimeter,
• $c_{\text{cu}}$ is the specific heat of  the copper,
• $T_i{,}_{\text{cal}}$ is the initial temperature of the calorimeter,

Heat gained by water and calorimeter is equal to the heat lost by the ice.

$Q_{\text{water}}+Q_{\text{cal}}=-\left(Q_{\text{ice}}+Q_{\text{melt}}+Q_{\text{ice-water}}\right)$ (6)

Substitute equation (1), (2), (3), (4), (5) in equation (6) and rewrite in terms of $T_f$ .

$\left\{\begin{array}{l}\left[m_{\text{water}}{c}_{\text{water}}\left(T_f-T_i{,}_{\text{water}}\right)+m_{\text{cal}}{c}_{\text{cu}}\left(T_f-T_i{,}_{\text{cal}}\right)\right]\\ =-\left[m_{\text{ice}}{c}_{\text{ice}}\left(T_f{,}_{\text{ice}}-T_i{,}_{\text{ice}}\right)+m_{\text{ice}}{L}_f+m_{\text{ice}}{c}_{\text{water}}\left(T_f-T_i{,}_{\text{ice-water}}\right)\right]\end{array}\right\}$

Substitute 40 g for $m_{\text{ice}}$, $2090\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{ice}}$, $0°\text{C}$ for $T_f{,}_{\text{ice}}$, $-78°\text{C}$ for $T_i{,}_{\text{ice}}$, $3.33\times {10}^5\text{J}/\text{kg}$ for $L_f$ , 560 g for $m_{\text{water}}$, $4186\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{water}}$, $25°\text{C}$ for $T_{i,\text{water}}$, $25°\text{C}$ for $T_{i,\text{cal}}$, $0°\text{C}$ for $T_i{,}_{\text{ice-water}}$, 80 g for $m_{\text{cal}}$, and $387\text{J}/\text{kg}\cdot \text{°C}$ for $c_{\text{cu}}$ to find $T_f$.

$\begin{array}{c}\left\{\begin{array}{l}\left(560\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)\left(T_f-\left(25°\text{C}\right)\right)\\ +\left(80\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(387\text{J}/\text{kg}\cdot \text{°C}\right)\left(T_f-\left(25°\text{C}\right)\right)\end{array}\right\}\\ =-\left\{\begin{array}{l}\left(40\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(2090\text{J}/\text{kg}\cdot \text{°C}\right)\left(0-\left(-78°\text{C}\right)\right)\\ +\left(40\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(3.33\times {10}^5\text{J}/\text{kg}\right)\\ +\left(40\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(4186\text{J}/\text{kg}\cdot \text{°C}\right)\left(T_f-\left(0°\text{C}\right)\right)\end{array}\right\}\\ \left\{\left(2344.16\text{J}/\text{°C}\right)T_f-58604\text{J}+\left(30.96\text{J}/\text{°C}\right)T_f-774\text{J}\right\}\\ =\left\{-6520.8\text{J}-13320\text{J}-\left(167.44\text{J}/\text{°C}\right)T_f\right\}\\ \left\{\left(2344.16\text{J}/\text{°C}\right)T_f+\left(30.96\text{J}/\text{°C}\right)T_f+\left(167.44\text{J}/\text{°C}\right)T_f\right\}\\ =\left\{58604\text{J}+774\text{J}-6520.8\text{J}-13320\text{J}\right\}\\ T_f=\frac{39537.2\text{J}}{2542.56\text{J}/\text{°C}}\\ \approx 16°\text{C}\end{array}$

The assumption of ice melts completely is correct. Since the final temperature is greater than $0°\text{C}$

Conclusion:

Therefore, the final temperature is $\text{16}°\text{C}$.