Chapter 14, Problem 52P

To determine

The mass of the ice added to cool the coffee.

Answer to Problem 52P

The mass of the ice added to cool the coffee is $36\text{g}$.

Explanation of Solution

Heat flows from coffee to the ice is taken as positive, when it enters the ice to melt it and warm the melt water, and as taken as negative when it leaves from the coffee.

The sum of the heat flow into the ice to cool the coffee is equal to zero.

$Q_{\text{ice}}+Q_{\text{C}}=0$ (I)

Here, $Q_{\text{ice}}$ is the amount of heat lost by the ice and $Q_{\text{C}}$ is the amount of heat gained by the coffee.

Write the equation for the amount of heat energy lost by the ice

$Q_{\text{ice}}=m_{\text{ice}}{c}_{\text{w}}\Delta T_{\text{ice}}$ (II)

Here, $m_{\text{w}}$ is the mass of the water, $c_{\text{w}}$ is the specific heat capacity of the water, and $\Delta T_{\text{ice}}$ is the change in temperature of ice cools.

Write the equation for latent heat of fusion (ice melts).

$Q_{\text{ice}}=m_{\text{ice}}{L}_{\text{f}}$ (III)

Here, $m_{\text{ice}}$ is the mass of the ice melts, and $L_{\text{f}}$ is the latent heat of fusion for ice melts.

Write the equation for the amount of heat energy gained by the cool coffee

$Q_{\text{C}}=m_{\text{C}}{c}_{\text{w}}\Delta T_{\text{C}}$ (IV)

Here, $m_{\text{C}}$ is the mass of the coffee, $c_{\text{w}}$ is the specific heat capacity of the water, and $\Delta T_{\text{C}}$ is the change in temperature of the coffee gets cooled.

Conclusion:

Substitute the equation (II), (III) and (IV) in equation (I) to solve for $m_{\text{ice}}$.

$\begin{array}{c}{m}_{\text{ice}}{L}_{\text{f}}+m_{\text{ice}}{c}_{\text{w}}\Delta T_{\text{ice}}+m_{\text{C}}{c}_{\text{w}}\Delta T_{\text{C}}=0\\ m_{\text{ice}}\left(L_{\text{f}}+c_{\text{w}}\Delta T_{\text{ice}}\right)+m_{\text{C}}{c}_{\text{w}}\Delta T_{\text{C}}=0\\ m_{\text{ice}}\left(L_{\text{f}}+c_{\text{w}}\Delta T_{\text{ice}}\right)=-m_{\text{C}}{c}_{\text{w}}\Delta T_{\text{C}}\\ m_{\text{ice}}=-\frac{m_{\text{C}}{c}_{\text{w}}\Delta T_{\text{C}}}{L_{\text{f}}+c_{\text{w}}\Delta T_{\text{ice}}}\end{array}$

Substitute $250\text{g}$ for $m_{\text{C}}$, $4186\text{J/kg}\cdot \text{K}$ for $c_{\text{w}}$, $80.0°\text{C}-60.0°\text{C}$ for $\Delta T_{\text{C}}$, $0°\text{C}-60.0°\text{C}$ for $\Delta T_{\text{ice}}$, and $333.7\text{kJ/kg}$ for $L_{\text{f}}$ in above equation.

$\begin{array}{c}{m}_{\text{ice}}=-\frac{\left(250\text{g}\right)\left(\frac{0.001\text{kg}}{1\text{g}}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left[\left(80.0\text{K}+273.15\text{K}\right)-\left(60.0\text{K}+273.15\text{K}\right)\right]}{\left(333.7\text{kJ/kg}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)+\left(4186\text{J/kg}\cdot \text{K}\right)\left[\left(0\text{K}+273.15\text{K}\right)-\left(60.0\text{K}+273.15\text{K}\right)\right]}\\ =-\frac{\left(0.250\text{kg}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(193.15\text{K}-213.15\text{K}\right)}{\left(333.7\times {10}^3\text{J/kg}\right)+\left(4186\text{J/kg}\cdot \text{K}\right)\left(273\text{K}-213.15\text{K}\right)}\\ =\frac{\left(0.250\text{kg}\right)\left(4186\text{J/kg}\cdot \text{K}\right)\left(20\text{K}\right)}{\left(333.7\times {10}^3\text{J/kg}\right)+\left(4186\text{J/kg}\cdot \text{K}\right)\left(60\text{K}\right)}\end{array}$

Solve the above relation for $m_{\text{ice}}$ in gram.

$\begin{array}{c}{m}_{\text{ice}}=\frac{20930\text{J}}{584860\text{J/kg}}\\ =0.036\text{kg}\left(\frac{1000\text{g}}{1\text{kg}}\right)\\ =36\text{g}\end{array}$

Therefore, the mass of the ice added to cool the coffee is $36\text{g}$.