Chapter 11, Problem 34P

(a)

To determine

Five thermal energy transfers that occur.

Explanation of Solution

Aluminum container at $20°\text{C}$ contains ethyl alcohol at $30°\text{C}$ and ice at $0°\text{C}$. Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at $-10°\text{C}$ increases in temperature and becomes Ice at $0°\text{C}$ , then phase change occurs from ice to cold water at $0°\text{C}$, then the temperature of the cold water at $0°\text{C}$ raises to a value T. Temperature of the Aluminum container at $20°\text{C}$ raises to a value T and ethyl alcohol at $30°\text{C}$ temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

(b)

To determine

Construct a table similar to table 11.5

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

$Q$	$m\left(\text{kg}\right)$	$c\left(\text{J}/\text{kg}\cdot \text{°C}\right)$	$L\left(\text{J}/\text{kg}\right)$	$T_f\left(°\text{C}\right)$	$T_i\left(°\text{C}\right)$	Expressions
$Q_{\text{ice}}$	1.00	2090	-	0	$-10.0$	$\begin{array}{l}{m}_{\text{ice}}{c}_{\text{ice}}\left[T_f-T_i\right]\\ =m_{\text{ice}}{c}_{\text{ice}}\left[0-\left(-10.0°\text{C}\right)\right]\end{array}$
$Q_{\text{melt}}$	1.00	-	$3.33\times {10}^5$	0	0	$m_{\text{ice}}{L}_f$
$Q_{\text{water}}$	1.00	4186	-	T	0	$\begin{array}{l}{m}_{\text{ice}}{c}_{\text{water}}\left[T_f-T_i\right]\\ =m_{\text{ice}}{c}_{\text{water}}\left[T_f-0\right]\end{array}$
$Q_{\text{Al}}$	0.500	900	-	T	20.0	$\begin{array}{l}{m}_{\text{Al}}{c}_{\text{Al}}\left[T_f-T_i\right]\\ =m_{\text{Al}}{c}_{\text{Al}}\left[T_f-20.0°\text{C}\right]\end{array}$
$Q_{\text{EAlc}}$	6.00	2430	-	T	30.0	$\begin{array}{l}{m}_{\text{EAlc}}{c}_{\text{EAlc}}\left[T_f-T_i\right]\\ =m_{\text{EAlc}}{c}_{\text{EAlc}}\left[T_f-30.0°\text{C}\right]\end{array}$

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

(c)

To determine

Conservation of energy equation.

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

$\left\{\begin{array}{l}{m}_{\text{ice}}{c}_{\text{ice}}\left[0-\left(-10.0°\text{C}\right)\right]+m_{\text{ice}}{L}_f+m_{\text{ice}}{c}_{\text{water}}\left[T_f-0\right]\\ +m_{\text{Al}}{c}_{\text{Al}}\left[T_f-20.0°\text{C}\right]+m_{\text{EAlc}}{c}_{\text{EAlc}}\left[T_f-30.0°\text{C}\right]=0\end{array}\right\}$

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

(d)

To determine

The final equilibrium temperature.

Answer to Problem 34P

The final equilibrium temperature is $T=4.81°\text{C}$

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is $20.0°\text{C}$, mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is $30.0°\text{C}$, mass of ice is 100 kg, initial temperature of ice is $-10.0°\text{C}$.

Formula to calculate the heat gained by cold water is,

$\left\{\begin{array}{l}{m}_{\text{ice}}{c}_{\text{ice}}\left[0-\left(-10.0°\text{C}\right)\right]+m_{\text{ice}}{L}_f+m_{\text{ice}}{c}_{\text{water}}\left[T_f-0\right]\\ +m_{\text{Al}}{c}_{\text{Al}}\left[T_f-20.0°\text{C}\right]+m_{\text{EAlc}}{c}_{\text{EAlc}}\left[T_f-30.0°\text{C}\right]=0\end{array}\right\}$

Rearrange in terms of $T_f$.

$\begin{array}{c}\left\{\begin{array}{l}{m}_{\text{ice}}{c}_{\text{ice}}\left(10.0°\text{C}\right)+m_{\text{ice}}{L}_f+\\ m_{\text{ice}}{c}_{\text{water}}{T}_f+m_{\text{Al}}{c}_{\text{Al}}{T}_f\\ -m_{\text{Al}}{c}_{\text{Al}}\left(20.0°\text{C}\right)+m_{\text{EAlc}}{c}_{\text{EAlc}}{T}_f\\ -m_{\text{EAlc}}{c}_{\text{EAlc}}\left(30.0°\text{C}\right)\end{array}\right\}=0\\ m_{\text{ice}}{c}_{\text{water}}{T}_f+m_{\text{Al}}{c}_{\text{Al}}{T}_f+m_{\text{EAlc}}{c}_{\text{EAlc}}{T}_f=\left\{\begin{array}{l}-m_{\text{ice}}{c}_{\text{ice}}\left(10.0°\text{C}\right)-m_{\text{ice}}{L}_f\\ +m_{\text{Al}}{c}_{\text{Al}}\left(20.0°\text{C}\right)+m_{\text{EAlc}}{c}_{\text{EAlc}}\left(30.0°\text{C}\right)\end{array}\right\}\\ T_f=\frac{\left\{\begin{array}{l}-m_{\text{ice}}\left[c_{\text{ice}}\left(10.0°\text{C}\right)+L_f\right]+m_{\text{Al}}{c}_{\text{Al}}\left(20.0°\text{C}\right)\\ +m_{\text{EAlc}}{c}_{\text{EAlc}}\left(30.0°\text{C}\right)\end{array}\right\}}{m_{\text{ice}}{c}_{\text{water}}+m_{\text{Al}}{c}_{\text{Al}}+m_{\text{EAlc}}{c}_{\text{EAlc}}}\end{array}$

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

$\begin{array}{c}{T}_f=\frac{\left\{\begin{array}{l}-\left(1.00\text{kg}\right)\left[\left(2090\text{J}/\text{kg°C}\right)\left(10.0°\text{C}\right)+\left(3.33\times {10}^5\text{J}/\text{kg}\right)\right]\\ +\left[\left(0.500\text{kg}\right)\left(900\text{J}/\text{kg°C}\right)\left(20.0°\text{C}\right)\right]\\ +\left[\left(6.00\text{kg}\right)\left(2430\text{J}/\text{kg°C}\right)\left(30.0°\text{C}\right)\right]\end{array}\right\}}{\left(\left(1.00\text{kg}\right)\left(4186\text{J}/\text{kg°C}\right)+\left(0.500\text{kg}\right)\left(900\text{J}/\text{kg°C}\right)+\left[\left(6.0\text{kg}\right)\left(2430\text{J}/\text{kg°C}\right)\right]\right)}\\ =\frac{92500}{19216}\\ =4.81°\text{C}\end{array}$

Conclusion:

Therefore, the final equilibrium temperature is $4.81°\text{C}$.