Chapter 15, Problem 110E

Interpretation Introduction

Interpretation:

The amount of energy must be removed from the tray to reduce the temperature from $\text{17 }°\text{C}$ to $\text{0 °C}$, freeze the water and drop the temperature of the tray and ice to $-\text{9 °C}$ is to be stated.

Concept introduction:

The amount of energy required to change the state of a substance is known as enthalpy. It is the different in the energy of final and initial state of a substance. The negative and positive sign of enthalpy indicates the energy released and energy absorbed, respectively, during the phase change.

Answer to Problem 110E

Amount of energy that must be removed from the tray to reduce the temperature from $\text{17 }°\text{C}$ to $-\text{9 °C}$ is $-\text{1}\text{.3 KJ}$.

Amount of energy that must be removed to cool the water from $\text{17 }°\text{C}$ to $\text{0 °C}$ is $-\text{29 KJ}$.

Amount of energy needed to freeze the water is $-\text{136 KJ}$.

Amount of energy removed to when temperature of ice is decreased from $\text{0 °C}$ to $-\text{9 °C}$ is $-\text{8 KJ}$.

Total energy removed or released is $-\text{174 KJ}$.

Explanation of Solution

The amount of heat required to decrease the temperature of aluminium from $\text{17 }°\text{C}$ to $-\text{9 °C}$ is calculated by the formula shown below.

${\text{q}}_{\text{1}}=\text{mc}\left({\text{T}}_2-{\text{T}}_1\right)$…(1)

Where,

• $\text{m}$ is the mass of the sample.

• $\text{c}$ is the specific heat of aluminium.

• ${\text{T}}_{\text{2}}$ is the final temperature.

• ${\text{T}}_{\text{1}}$ is the initial temperature.

The mass of the sample is $\left(\text{54}\text{.1g}\right)$.

The given final temperature is $-\text{9 °C}$.

The given initial temperature is $\text{17 }°\text{C}$.

The specific heat of aluminium is $0.90\text{J}/\text{g}\cdot °\text{C}$.

Substitute the values of mass of the sample, final temperature, initial temperature and specific heat of aluminium in equation (1).

$\begin{array}{rll}{\text{q}}_{\text{1}}& =& \text{54}\text{.1}\text{g}\times 0.90\text{J}/\text{g}\cdot °\text{C}\left(-\text{9 °C}-\left(\text{17 °C}\right)\right)\\ & =& -\text{1}\text{.3 KJ}\end{array}$

Negative sign shows that energy is released during decrease in temperature of aluminium from $\text{17}°\text{C}$ to $-9°\text{C}$.

Therefore, amount of heat required to decrease the temperature of aluminium from $\text{17 }°\text{C}$ to $-\text{9 °C}$ is $-1.3\times {10}^3\text{ KJ}$.

The amount of energy removed to decrease the temperature of water from $\text{17 }°\text{C}$ to $\text{0 °C}$ is shown below.

The given value of mass of water $\left(\text{m}\right)$ is $\text{408 g}$.

The given value of final temperature $\left({\text{T}}_{\text{2}}\right)$ is $\text{0 °C}$.

The given value of initial temperature $\left({\text{T}}_1\right)$ is $\text{17 }°\text{C}$.

The specific heat of water is $4.18\text{J}/\text{g}\cdot °\text{C}$.

Substitute the values of mass, final temperature, initial temperature and specific heat of water in equation (1).

$\begin{array}{rll}{\text{q}}_2& =& \text{408 g}\times \text{4}\text{.18 }\text{J}/\text{g}\times \text{°C}\left(\text{0 °C}-\left(\text{17 °C}\right)\right)\\ & =& -\text{29 KJ}\end{array}$

Therefore, amount of energy removed to decrease the temperature of cool water from $\text{17 }°\text{C}$ to $\text{0 °C}$ is $-29\times {10}^3\text{ KJ}$.

The amount of energy required for phase transformation is calculated by the formula shown below.

${\text{q}}_3=\text{m}\text{Δ}{\text{H}}_{\text{fus}}$…(2)

Where,

• $\text{Δ}{\text{H}}_{\text{fus}}$ is the heat of fusion.

The heat of fusion of water is $333\text{J}/\text{g}$.

Substitute the mass and heat of fusion in equation (2).

$\begin{array}{rll}{\text{q}}_3& =& 480\text{g}\times \left(-333\text{J}/\text{g}\right)\\ & =& -\text{136 KJ}\end{array}$

Negative sign indicates that energy is during the freezing process.

Therefore, amount of energy required for phase transformation is $-\text{136 KJ}$.

The amount of energy removed to decrease the temperature of water from $\text{0 °C}$ to $-\text{9 °C}$ is shown below.

The given value of mass of water $\left(\text{m}\right)$ is $\text{408 g}$.

The given value of final temperature $\left({\text{T}}_{\text{2}}\right)$ is $-\text{9 °C}$.

The given value of initial temperature $\left({\text{T}}_1\right)$ is $\text{0 °C}$.

The specific heat of ice is $2.06\text{J}/\text{g}\cdot °\text{C}$.

Substitute the mass, final temperature, initial temperature and specific heat of water in equation (1).

$\begin{array}{rll}{\text{q}}_4& =& \text{408 g}\times \text{2}\text{.06}\text{ J}/\text{g}\times \text{°C}\left(-\text{9 °C}-\left(\text{0 °C}\right)\right)\\ & =& -\text{8 KJ}\end{array}$

Therefore, amount of energy removed to decrease the temperature of ice from $\text{0 °C}$ to $-\text{9 °C}$ is $-\text{8 KJ}$.

The total amount of energy released is shown below.

$\begin{array}{rll}\text{q}& =& {\text{q}}_1+{\text{q}}_2+{\text{q}}_3+{\text{q}}_4\\ & =& \left(-1.3\right)+\left(-29\right)+(-\text{136})+\left(-8\right)\text{ KJ}\\ & =& -\text{174 KJ}\end{array}$

Conclusion

The amount of heat released to decrease the temperature of aluminium tray from $\text{17 °C}$ to $-\text{9 °C}$ is $-\text{1}\text{.3 KJ}$. Amount of energy released during decrease in temperature of water from $\text{17}°\text{C}$ to $0°\text{C}$ heat of fusion, and heat released when temperature is decreased from $\text{0 °C}$ to $-\text{9 °C}$ is $-1.3,-29,-\text{136 }$ and $\text{-8 KJ}$ respectively. Total heat released is $-\text{174 KJ}$.