Chapter 14, Problem 48P

(a)

To determine

The amount of ice need to be put in $0.250\text{kg}$ of water at $25.0\text{°C}$ to reduce the temperature of the water to $0\text{°C}$.

Answer to Problem 48P

The mass of ice need to be added is $74\text{g}$.

Explanation of Solution

The temperature of the ice is $-10.0\text{°C}$. The temperature of water is $25.0\text{°C}$. Mass of water is $0.250\text{kg}$. The final temperature is $0\text{°C}$.

The change in internal energy of the ice-water system is zero.

Write the equation for the first law of thermodynamic for the system.

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

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

The temperature of the water changes from $25.0\text{°C}$ to $0\text{°C}$.

Write the equation for the heat change of the water.

$Q_w=m_w{c}_w\left(T_f-T_{\text{wi}}\right)$ (II)

Here, $m_w$ is the mass of the water, $c_w$ is the specific heat of water, $T_f$ is the final temperature, $T_{\text{wi}}$ is the initial temperature of water.

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

Write the equation for the heat change of the ice.

$Q_i=m_i{c}_i\Delta T_1+m_{i}L$ (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.

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

$m_i{c}_i\Delta T_1+m_{i}L+m_w{c}_w\left(T_f-T_{\text{wi}}\right)=0$

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

$\begin{array}{c}{m}_i\left(c_i\Delta T_1+L\right)=-m_w{c}_w\left(T_f-T_{\text{wi}}\right)\\ m_i=\frac{-m_w{c}_w\left(T_f-T_{\text{wi}}\right)}{c_i\Delta T_1+L}\end{array}$ (IV)

Conclusion:

Substitute $0.250\text{kg}$ for $m_w$, $25.0\text{°C}$ for $T_{\text{wi}}$, $0.0\text{°C}$ for $T_f$, $0.0\text{°C}-\left(-10.0\text{°C}\right)$ for $\Delta T_1$, $4.186\text{kJ/kg}\cdot \text{K}$ for $c_w$, $2.1\text{kJ/kg}\cdot \text{K}$ for $c_i$, $333.7\text{kJ/kg}$ for $L$.

$\begin{array}{c}{m}_i=\frac{-\left(0.250\text{kg}\right)\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(0.0\text{°C}-25.0\text{°C}\right)}{\left(2.1\text{kJ/kg}\cdot \text{K}\right)\left(0.0\text{°C}-\left(-10.0\text{°C}\right)\right)+333.7\text{kJ/kg}}\\ =74\text{g}\end{array}$

The mass of ice need to be added is $74\text{g}$.

(b)

To determine

The final temperature if half of the ice as section (a) is added to the water.

Answer to Problem 48P

The final temperature is $\text{11}\text{°C}$.

Explanation of Solution

The temperature of the ice is $-10.0\text{°C}$. The temperature of water is $25.0\text{°C}$. Mass of water is $0.250\text{kg}$. The final temperature is $0\text{°C}$.

The change in internal energy of the ice-water system is zero.

Write the equation for the first law of thermodynamic for the system.

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

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

The temperature of the water changes from $25.0\text{°C}$ to $0\text{°C}$.

Write the equation for the heat change of the tea.

$Q_w=m_w{c}_w\left(T_f-T_{\text{wi}}\right)$ (II)

Here, $m_w$ is the mass of the water, $c_w$ is the specific heat of water, $T_f$ is the final temperature, $T_{\text{wi}}$ is the initial temperature of water.

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 changes from $0.0\text{°C}$ to the final temperature.

Write the equation for the heat change of the ice.

$Q_i=m_i{c}_i\left(T_{ii}-\text{0}\text{°C}\right)+m_{i}L+m_i{c}_w\left(T-\text{0}\text{°C}\right)$ (III)

Here, $m_i$ is the mass of ice, $c_i$ is the specific heat of ice, $T_{ii}$ is the initial temperature of the ice, $L$ is the latent heat of fusion of water.

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

$m_i{c}_i\left(T_{ii}-\text{0}\text{°C}\right)+m_{i}L+m_i{c}_w\left(T-\text{0}\text{°C}\right)+m_w{c}_w\left(T_f-T_{\text{wi}}\right)=0$

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

$\begin{array}{c}{T}_f{c}_w\left(m_w+m_i\right)=-m_i{c}_i\left(T_{ii}-\text{0}\text{°C}\right)-m_{i}L+m_i{c}_w\left(\text{0}\text{°C}\right)+m_w{c}_w{T}_{\text{wi}}\\ T_f=\frac{-m_i{c}_i\left(T_{ii}-\text{0}\text{°C}\right)-m_{i}L+m_i{c}_w\left(\text{0}\text{°C}\right)+m_w{c}_w{T}_{\text{wi}}}{c_w\left(m_w+m_i\right)}\end{array}$

Conclusion:

Substitute $0.250\text{kg}$ for $m_w$, $25.0\text{°C}$ for $T_{\text{wi}}$, $74\text{g}/2$ for $m_i$, $4.186\text{kJ/kg}\cdot \text{K}$ for $c_w$, $2.1\text{kJ/kg}\cdot \text{K}$ for $c_i$, $333.7\text{kJ/kg}$ for $L$, $-10.0\text{°C}$ for $T_{ii}$.

$\begin{array}{c}{T}_f=\frac{\left[\begin{array}{l}-\left(74\text{g}/2\right)\left(2.1\text{kJ/kg}\cdot \text{K}\right)\left(-10.0\text{°C}-\text{0}\text{°C}\right)-\left(74\text{g}/2\right)\left(333.7\text{kJ/kg}\right)\\ +\left(74\text{g}/2\right)\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(\text{0}\text{°C}\right)+\left(0.250\text{kg}\right)\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(25.0\text{°C}\right)\end{array}\right]}{\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(0.250\text{kg}+74\text{g}/2\right)}\\ =\frac{\left[\begin{array}{l}-\left(74\text{g}/2\right)\left(2.1\text{kJ/kg}\cdot \text{K}\right)\left(-10.0+273\text{K}-\text{0}+273\text{K}\right)-\left(74\text{g}/2\right)\left(333.7\text{kJ/kg}\right)\\ +\left(74\text{g}/2\right)\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(\text{0}+273\text{K}\right)+\left(0.250\text{kg}\right)\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(25.0+273\text{K}\right)\end{array}\right]}{\left(4.186\text{kJ/kg}\cdot \text{K}\right)\left(0.250\text{kg}+74\text{g}/2\right)}\\ =284\text{K}\\ =284-273\text{°C}\\ \text{=11}\text{°C}\end{array}$

The final temperature is $\text{11}\text{°C}$.