Chapter 18, Problem 41P

(a)

To determine

The final temperature of the system.

Answer to Problem 41P

The final temperature of the system is $5.26°\text{C}$ .

Explanation of Solution

Given:

The mass of aluminium calorimeter is $m_{\text{Al calorimeter}}=200\text{g}$

The mass of water is $m_{\text{water}}=600\text{g}$

The mass of ice piece is $m_{\text{ice}}=100\text{g}$

Formula used:

The expression for the energy available to melt the ice is given as,

  $Q_{\text{available}}=m_{\text{water}}{c}_{\text{water}}\Delta T_{\text{water}}+m_{\text{Al calorimeter}}{c}_{\text{Al calorimeter}}\Delta T_{\text{Al calorimeter}}$

The expression for the energy required to warm and melt the ice is given as,

  $Q_{\text{warm and melt ice}}=m_{\text{ice}}{c}_{\text{ice}}\Delta T_{\text{ice}}+m_{\text{ice}}{L}_f$

Here, $L_f$ is the latent heat of fusion.

The expression for the law of conservation of energy is given as,

  $\begin{array}{l}\sum Q_i=Q_{\text{cool calorimeter}}+Q_{\text{cool warm water}}+Q_{\text{warm and melt ice}}+Q_{\text{warm ice water}}=0\\ \left[\begin{array}{l}{m}_{\text{cool calorimeter }}{c}_{\text{Al calorimeter}}\left(t_f-20°\text{C}\right)+m_{\text{water}}{c}_{\text{water}}\left(t_f-\left(-20°\text{C}\right)\right)+\\ m_{\text{ice}}{c}_{\text{water}}\left(t_f-0°\text{C}\right)+35.45\text{kJ}\end{array}\right]=0\end{array}$

Calculation:

The energy available to melt the ice is calculated as,

  $\begin{array}{c}{Q}_{\text{available}}=m_{\text{water}}{c}_{\text{water}}\Delta T_{\text{water}}+m_{\text{Al calorimeter}}{c}_{\text{Al calorimeter}}\Delta T_{\text{Al calorimeter}}\\ =\left[\begin{array}{l}600\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(4.184\text{kJ}/\text{kg}\cdot \text{K}\right)+\\ 200\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(0.900\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0°\text{C}-20°\text{C}\right)\end{array}\right]\\ =53.81\text{kJ}\end{array}$

The energy required to warm and melt the ice is calculated as,

  $\begin{array}{c}{Q}_{\text{warm and melt ice}}=m_{\text{ice}}{c}_{\text{ice}}\Delta T_{\text{ice}}+m_{\text{ice}}{L}_f\\ =\left[\begin{array}{l}100\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(2.05\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(0°\text{C}+20°\text{C}\right)+\\ 100\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times 333.5\text{kJ}/\text{kg}\end{array}\right]\\ =37.45\text{kJ}\end{array}$

By the law of conservation of energy, the final temperature is calculated as,

  $\begin{array}{l}\left[\begin{array}{l}{m}_{\text{cool calorimeter }}{c}_{\text{Al calorimeter}}\left(t_f-20°\text{C}\right)+m_{\text{water}}{c}_{\text{water}}\left(t_f-\left(-20°\text{C}\right)\right)\\ +m_{\text{ice}}{c}_{\text{water}}\left(t_f-0°\text{C}\right)+35.45\text{kJ}\end{array}\right]=0\\ \left[\begin{array}{l}200\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(0.900\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(t_f-20°\text{C}\right)+\\ 600\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(4.184\text{kJ}/\text{kg}\cdot \text{K}\right)\left(t_f+20°\text{C}\right)\\ +100\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(4.184\text{kJ}/\text{kg}\cdot \text{K}\right)\left(t_f-0°\text{C}\right)+35.45\text{kJ}\end{array}\right]=0\\ t_f=5.26°\text{C}\end{array}$

Conclusion:

Therefore, the final temperature of the system is $5.26°\text{C}$ .

(b)

To determine

The ice left in the system after reaches the equilibrium state.

Answer to Problem 41P

The ice left in the system after reaches the equilibrium state is $200\text{g}$ .

Explanation of Solution

Given:

The mass of ice piece is $m_{\text{ice}}=200\text{g}$

Formula used:

The expression for the mass of ice is given as,

  $m_{\text{ice remaining}}=m_{\text{ice initially}}-m_{\text{ice melted}}=m_{\text{ice initially}}-\frac{Q_{\text{avaialble}}-Q_{\text{warm ice}}}{L_f}$

The value of $Q_{\text{warm ice}}$ can be calculated as,

  $Q_{\text{warm ice}}=m_{\text{ice}}{c}_{\text{ice}}\Delta T_{\text{ice}}$

The value of $Q_{\text{available}}$ can be calculated as,

  $Q_{\text{available}}=m_{\text{water}}{c}_{\text{water }}\Delta T_{\text{water}}+m_{\text{calorimeter}}{c}_{\text{calorimeter}}\Delta T_{\text{water}}$

Calculation:

The value of $Q_{\text{warm ice}}$ can be calculated as,

  $\begin{array}{c}{Q}_{\text{warm ice}}=m_{\text{ice}}{c}_{\text{ice}}\Delta T_{\text{ice}}\\ =200\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(2.02\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(0°\text{C}+20°\text{C}\right)\\ =8.080\text{kJ}\end{array}$

The value of $Q_{\text{available}}$ can be calculated as,

  $\begin{array}{c}{Q}_{\text{available}}=m_{\text{water}}{c}_{\text{water }}\Delta T_{\text{water}}+m_{\text{calorimeter}}{c}_{\text{calorimeter}}\Delta T_{\text{water}}\\ =\left[\begin{array}{l}700\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\times \left(4.184\text{kJ}/\text{kg}\cdot \text{K}\right)+\\ \left(0.2\text{kg}\right)\times \left(0.900\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\left(5.262°\text{C}-0°\text{C}\right)\\ =16.359\text{kJ}\end{array}$

The remaining mass of ice is calculated as

  $\begin{array}{c}{m}_{\text{ice remaining}}=m_{\text{ice initially}}-\frac{Q_{\text{avaialble}}-Q_{\text{warm ice}}}{L_f}\\ =\left(200\text{g}\right)-\frac{\left(16.359\text{kJ}\right)-\left(8.080\text{kJ}\right)}{\left(333.5\text{kJ}/\text{kg}\right)}\\ =200\text{g}\end{array}$

Conclusion:

Therefore, the ice left in the system after reaches the equilibrium state is $200\text{g}$ .

(c)

To determine

The change if both pieces of ice were mix at the same time.

Explanation of Solution

Introduction:

Two bodies are said to be in thermal equilibrium if they are in a close connection that permits either to obtain energy from the other. Still, nevertheless, no net energy is transferred between them. During the process of reaching thermal equilibrium, heat, which is a form of energy, is transferred between the objects.

If both pieces of ice were mix at the same time, then there is no change as the initial and final conditions in equilibrium state are identical.

Conclusion:

Therefore, there is no change at the time of mixing of both ice pieces.