Chapter 14, Problem 41P

To determine

How much ice should be added to the glass of tea to cool it to a temperature of $10.0\text{°C}$.

Answer to Problem 41P

The mass of ice need to be added is $179\text{g}$.

Explanation of Solution

The temperature of the ice is $-10.0\text{°C}$. The volume of tea is $2.00\times {10}^{-4}{\text{m}}^{\text{3}}$. The initial temperature of hot tea is $95.0\text{°C}$. The final temperature of tea is $10.0\text{°C}$.

Since the container is insulated, the change in internal energy of the tea-ice system is zero.

Write the equation for the first law of thermodynamic for the system of ball and water.

$\begin{array}{c}{Q}_t+Q_i=\Delta U\\ =0\end{array}$ (I)

Here, $\Delta U$ is the change in internal energy, $Q_t$ is the heat of the tea, $Q_i$ is the ice.

The temperature of the tea changes from $95.0\text{°C}$ to $10.0\text{°C}$.

Write the equation for the heat change of the tea.

$Q_t={\rho }_w{V}_t{c}_w\left(T_f-T_{\text{ti}}\right)$ (II)

Here, ${\rho }_w$ is the density of the water, $V_t$ is the volume of tea, $c_w$ is the specific heat of water, $T_f$ is the final temperature, $T_{\text{ti}}$ is the initial temperature of tea.

The temperature of the ice first changes from $-10.0\text{°C}$ to $0.0\text{°C}$, then ice turns to water at $0.0\text{°C}$, then the temperature of the water changed from $0.0\text{°C}$ to $10.0\text{°C}$.

Write the equation for the heat change of the ice.

$Q_i=m_i{c}_i\Delta T_1+m_{i}L+m_i{c}_w\Delta T_2$ (III)

Here, $m_i$ is the mass of ice, $c_i$ is the specific heat of ice, $\Delta T_1$ is the change in temperature from $-10.0\text{°C}$ to $0.0\text{°C}$, $L$ is the latent heat of fusion of water, $\Delta T_2$ is the temperature change $0.0\text{°C}$ to $10.0\text{°C}$.

Substitute equation (II) and (III) in equation (I).

$m_i{c}_i\Delta T_1+m_{i}L+m_i{c}_w\Delta T_2+{\rho }_w{V}_t{c}_w\left(T_f-T_{\text{ti}}\right)=0$

Re-write the above equation to get an expression for $m_i$.

$\begin{array}{c}{m}_i\left(c_i\Delta T_1+L+c_w\Delta T_2\right)=-{\rho }_w{V}_t{c}_w\left(T_f-T_{\text{ti}}\right)\\ m_i=\frac{-{\rho }_w{V}_t{c}_w\left(T_f-T_{\text{ti}}\right)}{c_i\Delta T_1+L+c_w\Delta T_2}\end{array}$ (IV)

Conclusion:

Substitute $2.00\times {10}^{-4}{\text{m}}^{\text{3}}$ for $V_t$, $95.0\text{°C}$ for $T_{\text{ti}}$, $10.0\text{°C}$ for $T_f$, $0.0\text{°C}-\left(-10.0\text{°C}\right)$ for $\Delta T_1$, $10.0\text{°C}-0.0\text{°C}$ for $\Delta T_2$, $1000{\text{kg/m}}^{\text{3}}$ for ${\rho }_w$, $4.186\text{kJ/kg}\cdot \text{K}$ for $c_w$, $2.1\text{kJ/kg}\cdot \text{K}$ for $c_i$, $333.7\text{kJ/kg}$ for $L$.

$\begin{array}{c}{m}_i=\frac{-\left(1000{\text{kg/m}}^{\text{3}}\right)\left(2.00\times {10}^{-4}{\text{m}}^{\text{3}}\right)\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(10.0\text{°C}-95.0\text{°C}\right)}{\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(0.0\text{°C}-\left(-10.0\text{°C}\right)\right)+333.7\text{kJ/kg}+\left(10.0\text{°C}-0.0\text{°C}\right)\left(2.1\text{kJ/kg}\cdot \text{K}\right)}\\ =179\text{g}\end{array}$

The mass of ice need to be added is $179\text{g}$.