Chapter 22, Problem 22.70AP

A biology laboratory is maintained at a constant temperature of 7.00ºC by an air conditioner, which is vented to the air outside. On a typical hot summer day, the outside temperature is 27.0ºC and the air-conditioning unit emits energy to the outside at a rate of 10.0 kW. Model the unit as having a coefficient of performance (COP) equal to 40.0% of the COP of an ideal Carnot device. (a) At what rate does the air conditioner remove energy from the laboratory? (b) Calculate the power required for the work input. (c) Find the change in entropy of the Universe produced by the air conditioner in 1.00 h. (d) What If? The outside temperature increases to 32.0ºC. Find the fractional change in the COP of the air conditioner.

To determine

The rate of removal of energy by the air conditioner.

Answer to Problem 22.70AP

The rate of removal of energy by the air conditioner is $8.48\text{kW}$.

Explanation of Solution

Given info: The temperature of laboratory is $7.00°\text{C}$. The temperature outside the laboratory is $27.0°\text{C}$. The rate at which air conditioner emits energy outside is $10.0\text{kW}$. The C.O.P of model is $40.0\%$.

The formula for Carnot efficiency of cooling is,

$E_{\text{c}}=\frac{T_{\text{l}}}{T_{\text{o}}-T_{\text{l}}}$

Here,

$E_{\text{c}}$ is Carnot efficiency of cooling.

$T_{\text{l}}$ is room temperature.

$T_{\text{o}}$ is temperature outside the room.

Substitute $7.00°\text{C}$ for $T_{\text{l}}$ and $27.0°\text{C}$ for $T_{\text{o}}$ in the above expression.

$E_{\text{c}}=\frac{7.00°\text{C}}{27.0°\text{C}-7.00°\text{C}}$

Solve the above expression for $E_{\text{c}}$,

$\begin{array}{c}{E}_{\text{c}}=\frac{\left(7.00°+273°\right)\text{K}}{\left(27.0°+273°\right)\text{C}-\left(7.00°+273°\right)\text{K}}\\ =\frac{280.0\text{K}}{300.0\text{K}-280.0\text{K}}\\ =14\end{array}$

The coefficient of performance at $27°\text{C}$  is equals to $40\%$ of $E_{\text{c}}$.

${\text{COP}}_{27°\text{C}}\text{=40\%}\left(E_{\text{c}}\right)$,

Substitute $14$ for $E_{\text{c}}$ in the above expression.

$\begin{array}{c}{\text{COP}}_{27°\text{C}}\text{=40\%}\left(14\right)\\ =\text{5}\text{.6}\end{array}$

The rate of emission of energy outside is the sum of rate of removal of energy and input work required to do so.

$Q_{\text{in}}+W=Q_{\text{out}}$

Rearrange the above expression for $Q_{\text{in}}$,

$Q_{\text{in}}=Q_{\text{out}}-W$ (1)

The formula to coefficient of performance is,

$\frac{Q_{\text{in}}}{W}=C.O.P_{27°\text{C}}$

Here,

$C.O.P_{27°\text{C}}$ is coefficient of performance.

$Q_{\text{in}}$ is the rate of removal of energy.

$W$ is input work.

Substitute $Q_{\text{out}}-W$ for $Q_{\text{in}}$ in the above expression.

$\frac{Q_{\text{out}}-W}{W}=C.O.P_{27°\text{C}}$

Solve the above expression for $W$,

$\frac{Q_{\text{out}}}{W}-1=C.O.P_{27°\text{C}}$

Substitute $10.0\text{kW}$ for $Q_{\text{out}}$ and $\text{5}\text{.6}$ for $C.O.P_{27°\text{C}}$ in the above expression.

$\begin{array}{c}\frac{10}{W}-1=5.6\\ W=1.515\text{KW}\end{array}$ (2)

Substitute $1.515\text{kW}$ for $W$ and $10.0\text{kW}$ for $Q_{\text{out}}$ in equation (1).

$\begin{array}{c}{Q}_{\text{in}}=\left(10.0\text{KW}\right)-\left(1.515\text{KW}\right)\\ =8.48\text{kW}\end{array}$

Conclusion:

Therefore, the rate of removal of energy by the air conditioner is $8.48\text{kW}$.

To determine

The power required for the input work.

Answer to Problem 22.70AP

The power required for the input work is $1.52\text{kW}$.

Explanation of Solution

Given info: The temperature of laboratory is $7.00°\text{C}$. The temperature outside the laboratory is $27.0°\text{C}$. The rate at which air conditioner emits energy outside is $10.0\text{kW}$. The C.O.P of model is $40.0\%$.

As calculated in equation (2) of the above part,

$\begin{array}{c}W=1.515\text{kW}\\ \approx \text{1}\text{.52}\text{kw}\end{array}$

So the work input is $1.52\text{kW}$.

Conclusion:

Therefore, the work input required for the input work is $1.52\text{kW}$.

To determine

The entropy change of universe produced by air conditioner in $1.00\text{h}$.

Answer to Problem 22.70AP

The entropy change of universe produced by the air conditioner in $1.00\text{h}$ is $1.09\times {10}^4\text{J}/\text{K}$.

Explanation of Solution

Given info: The temperature of laboratory is $7.00°\text{C}$. The temperature outside the laboratory is $27.0°\text{C}$. The rate at which air conditioner emits energy outside is $10.0\text{kW}$. The C.O.P of model is $40.0\%$. The time for the air conditioner is $1.00\text{h}$.

The formula to calculate change in entropy is,

$\Delta S=\left(\frac{Q_{\text{out}}}{T_o}-\frac{Q_{\text{in}}}{T_{\text{l}}}\right)t$

Substitute $10.0\text{kW}$ for $Q_{\text{out}}$, $300°\text{K}$ for $T_{\text{o}}$, $8.485\text{kW}$ for $Q_{\text{in}}$, $280°\text{K}$ for $T_{\text{l}}$ and $1.00\text{h}$ for $t$ in the above expression.

$\Delta S=\left(\frac{10.0\text{kW}}{300°\text{K}}-\frac{8.485\text{kW}}{280°\text{K}}\right)\left(1.00\text{h}\right)$

Solve the above expression for $\Delta S$.

$\begin{array}{c}\Delta S=\left(\frac{10.0\text{kW}\times \left(\frac{{10}^3\text{W}}{1\text{kW}}\right)}{300°\text{K}}-\frac{8.485\text{kW}\times \left(\frac{{10}^3\text{W}}{1\text{kW}}\right)}{280°\text{K}}\right)\left(1.00\text{h}\times \left(\frac{3600\text{s}}{1.00\text{h}}\right)\right)\\ =10907.14\\ =1.09\times {10}^4\text{J}/\text{K}\end{array}$

Conclusion:

Therefore, the entropy change of universe produced by the air conditioner in $1.00\text{h}$ is $1.09\times {10}^4\text{J}/\text{K}$.

To determine

The fractional change in $COP$ of the air conditioner.

Answer to Problem 22.70AP

The fractional change in $COP$ of the air conditioner is $20\%$.

Explanation of Solution

Given info: The temperature of laboratory is $7.00°\text{C}$. The temperature outside the laboratory is $27.0°\text{C}$. The rate at which air conditioner emits energy outside is $10.0\text{kW}$. The C.O.P of model is $40.0\%$. The time for the air conditioner is $1.00\text{h}$. The temperature outside increases to $32.0°\text{C}$.

The formula to calculate $COP$ is,

$CO{P}_{32°\text{C}}=\frac{T_{\text{l}}}{T_{\text{o}}-T_{\text{l}}}$

Here,

$CO{P}_{32°\text{C}}$ is coefficient of performance at $32°\text{C}$.

Substitute $7.00°\text{C}$ for $T_{\text{l}}$ and $32.0°\text{C}$ for $T_{\text{o}}$ in the above expression.

$\begin{array}{c}CO{P}_{32°\text{C}}=\frac{7.00°\text{C}}{32.0°\text{C}-7.00°\text{C}}\\ =\frac{\left(273°+7.0°\right)\text{K}}{\left(273°+32.0°\right)\text{K}-\left(273°+7.0°\right)\text{K}}\\ =\frac{280°\text{K}}{305°\text{K}-\text{280}°\text{K}}\\ =11.2\end{array}$

The formula for the percentage is,

$CO{P}_{\text{change}}=\left(\frac{CO{P}_{27°\text{C}}-CO{P}_{32°\text{C}}}{CO{P}_{27°\text{C}}}\right)\left(100\right)$

Here,

$CO{P}_{\text{change}}$ is the change in coefficient of performance as compared to the previous coefficient of performance.

$CO{P}_{32°\text{C}}$ is coefficient of performance at $32°\text{C}$.

Substitute $14$ for $CO{P}_{27°\text{C}}$ and $11.2$ for $CO{P}_{32°\text{C}}$ in the above expression.

$\begin{array}{c}CO{P}_{\text{change}}=\left(\frac{14-11.2}{14}\right)\left(100\right)\\ =20\%\end{array}$

Conclusion:

Therefore, the fractional change in $COP$ of the air conditioner is $20\%$.