Chapter 22, Problem 70AP

(a)

To determine

The rate at which air conditioner removes energy from the laboratory.

Answer to Problem 70AP

The rate at which air conditioner removes energy from the laboratory is $8.48\text{kW}$ .

Explanation of Solution

An air conditioner removes energy from the cold reservoir to heat reservoir. Ideal air conditioner is the one which work in Carnot cycle.

Write the expression for the coefficient of performance of ideal air conditioner.

  ${\left(\text{COP}\right)}_{\text{Carnot}}=\frac{T_c}{T_h-T_c}$                                                                                                  (I)

Here, ${\left(\text{COP}\right)}_{\text{Canot}}$ is the coefficient of performance of ideal air conditioner, $T_c$ is the temperature of cold reservoir and $T_h$ is the temperature of hot reservoir.

Write the expression for the coefficient of performance of practical air refrigerator.

  $\text{COP}=\frac{\left|Q_c\right|}{\left|Q_h\right|-\left|Q_c\right|}$                                                                                                      (II)

Here, $\text{COP}$ is the coefficient of performance of air refrigerator, $\left|Q_c\right|$ is the energy by heat extracted from the cold reservoir and $\left|Q_h\right|$ is the energy by heat expelled at hot reservoir.

Write the expression to convert temperature in degree Celsius into Kelvin scale.

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$                                                                                              (III)

Here, $T\left(\text{K}\right)$ is the temperature in Kelvin scale and $T\left(°\text{C}\right)$ is the temperature in degree Celsius.

The coefficient of performance of the given air conditioner is $40.0\%$ that of the COP of an ideal Carnot device.

Write the relation between $\text{COP}$ and ${\left(\text{COP}\right)}_{\text{Canot}}$.

  $\text{COP}=40.0\%{\left(\text{COP}\right)}_{\text{Canot}}$                                                                                       (IV)

Conclusion:

The temperature of laboratory is $7.00°\text{C}$ and outside temperature is $27.0°\text{C}$.

Substitute $7.00°\text{C}$ for $T\left(°\text{C}\right)$ in equation (III) to get $T\left(\text{K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=7.00°\text{C}+273\\ =280\text{K}\end{array}$

Substitute $27.0°\text{C}$ for $T\left(°\text{C}\right)$ in equation (III) to get $T\left(\text{K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=27.0°\text{C}+273\\ =300\text{K}\end{array}$

Substitute $280\text{K}$ for $T_c$ and $300\text{K}$ for $T_h$ in equation (I) to get ${\left(\text{COP}\right)}_{\text{Carnot}}$.

  $\begin{array}{c}{\left(\text{COP}\right)}_{\text{Carnot}}=\frac{280\text{K}}{300\text{K}-280\text{K}}\\ =\frac{280\text{K}}{20.0\text{K}}\\ =14.0\end{array}$

Substitute $14.0$ for ${\left(\text{COP}\right)}_{\text{Carnot}}$ in (IV) to get $\text{COP}$.

  $\begin{array}{c}\text{COP}=40.0\%\times \frac{1}{100}\times 14.0\\ =5.60\end{array}$

Divide numerator and denominator of right hand side of equation (II) by $\Delta t$.

  $\text{COP}=\frac{\frac{\left|Q_c\right|}{\Delta t}}{\frac{\left|Q_h\right|}{\Delta t}-\frac{\left|Q_c\right|}{\Delta t}}$

It is given that air conditioner emits energy to the outside at the rate of $10.0\text{kW}$.

Substitute $5.60$ for $\text{COP}$ and $10.0\text{kW}$ for $\frac{\left|Q_h\right|}{\Delta t}$ in above equation to get $\frac{\left|Q_c\right|}{\Delta t}$.

  $\begin{array}{c}5.60=\frac{\frac{\left|Q_c\right|}{\Delta t}}{10.0\text{kW}-\frac{\left|Q_c\right|}{\Delta t}}\\ 5.60\left(10.0\text{kW}\right)-5.60\left(\frac{\left|Q_c\right|}{\Delta t}\right)=\frac{\left|Q_c\right|}{\Delta t}\\ 5.60\left(10.0\text{kW}\right)=\frac{\left|Q_c\right|}{\Delta t}+5.60\left(\frac{\left|Q_c\right|}{\Delta t}\right)\\ 5.60\left(10.0\text{kW}\right)=\frac{\left|Q_c\right|}{\Delta t}\left(6.60\right)\end{array}$

Rearrange above equation to get $\frac{\left|Q_c\right|}{\Delta t}$

  $\begin{array}{c}\frac{\left|Q_c\right|}{\Delta t}=\frac{5.60\left(10.0\text{kW}\right)}{\left(6.60\right)}\\ =8.48\end{array}$

Therefore, the rate at which air conditioner removes energy from the laboratory is $8.48\text{kW}$ .

(b)

To determine

The power required for the input work.

Answer to Problem 70AP

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

Explanation of Solution

Write the expression for the work done.

  $W_{\text{eng}}=\left|Q_h\right|-\left|Q_c\right|$                                                                                                       (V)

Here, $W_{\text{eng}}$ is the work done.

Divide both sides of the equation by $\Delta t$ to get expression for the power required for the input work.

  $\frac{W_{\text{eng}}}{\Delta t}=\frac{\left|Q_h\right|}{\Delta t}-\frac{\left|Q_c\right|}{\Delta t}$                                                                                                   (VI)

Conclusion:

Substitute $10.0\text{kW}$ for $\frac{\left|Q_h\right|}{\Delta t}$ and $8.48\text{kW}$ for $\frac{\left|Q_c\right|}{\Delta t}$ in above equation to get power required for the input work.

  $\begin{array}{c}\frac{W_{\text{eng}}}{\Delta t}=10.0\text{kW}-8.48\text{kW}\\ \text{=1}\text{.52}\text{kW}\end{array}$

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

(c)

To determine

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

Answer to Problem 70AP

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

Explanation of Solution

The entropy of the working substance does not change, since air conditioner operates in a cycle.

Write the expression for the increase in entropy of the hot reservoir.

  $\Delta S_1=\frac{\left|Q_h\right|}{T_h}$                                                                                                             (VII)

Here, $\Delta S_1$ is the increase in entropy of hot reservoir.

Write the expression for the decrease in entropy of the cold reservoir.

  $\Delta S_2=-\frac{\left|Q_c\right|}{T_c}$                                                                                                         (VIII)

Here, $\Delta S_2$ is the decrease in entropy of cold reservoir.

Write the expression for the change in entropy produced by the air conditioner in $1.00\text{h}$.

  $\Delta S=\Delta S_1+\Delta S_2$                                                                                                       (IX)

Here, $\Delta S$ is the change in entropy produced by the air conditioner in $1.00\text{h}$.

Use equation (VII) and (VIII) in equation (IX) to get $\Delta S$.

  $\Delta S=\frac{\left|Q_h\right|}{T_h}-\frac{\left|Q_c\right|}{T_c}$                                                                                                       (X)

Divide and multiply $\Delta t$ in numerator of right hand side of the equation to get $\Delta S$.

  $\Delta S=\frac{\frac{\left|Q_h\right|}{\Delta t}\Delta t}{T_h}-\frac{\frac{\left|Q_c\right|}{\Delta t}\Delta t}{T_c}$                                                                                            (XI)

Conclusion:

Substitute $10.0\text{kW}$ for $\frac{\left|Q_h\right|}{\Delta t}$ and $8.48\text{kW}$ for $\frac{\left|Q_c\right|}{\Delta t}$, $1\text{h}$ for $\Delta t$, $300\text{K}$ for $T_h$ and $280\text{K}$ for $T_c$ in equation (XI) to get $\Delta S$.

  $\begin{array}{c}\Delta S=\frac{\left(10.0\text{kW}\times \frac{1000\text{W}}{1\text{kW}}\right)\left(1\text{h}\times \frac{\text{3600}\text{s}}{1\text{h}}\right)}{300\text{K}}-\frac{\left(8.48\text{kW}\times \frac{1000\text{W}}{1\text{kW}}\right)\left(1\text{h}\times \frac{\text{3600}\text{s}}{1\text{h}}\right)}{280\text{K}}\\ =1.20\times {10}^5\text{J}/\text{K}-1.09\times {10}^5\text{J}/\text{K}\\ =1.09\times {10}^4\text{J}/\text{K}\end{array}$

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

(d)

To determine

The fractional change in the COP of the air conditioner.

Answer to Problem 70AP

The fractional change of the COP of the air conditioner is to drop by $20.0\%$.

Explanation of Solution

It is given that outside temperature is raised to $32.0°\text{C}$ from $27.0°\text{C}$.

Use equation (I) to get ideal coefficient of performance and equation (IV) to get actual coefficient of performance.

Use equation (III) to convert temperature in degree Celsius into Kelvin scale.

Write the expression for the fraction al change.

  $\text{fractional change}=\frac{\text{final COP}}{\text{intial COP}}$                                                                             (XII)

Conclusion:

Substitute $32.0°\text{C}$ for $T\left(°\text{C}\right)$ in equation (III) to convert in to Kelvin scale.

  $\begin{array}{c}T\left(\text{K}\right)=32.0°\text{C}+273\\ =305\text{K}\end{array}$

Substitute $305\text{K}$ for $T_h$, $280\text{K}$ for $T_c$ in equation (I) to get ${\left(\text{COP}\right)}_{\text{Carnot}}$.

  $\begin{array}{c}{\left(\text{COP}\right)}_{\text{Carnot}}=\frac{280\text{K}}{305\text{K}-280\text{K}}\\ =\frac{280\text{K}}{25\text{K}}\\ =11.2\end{array}$

Substitute $11.2$ for ${\left(\text{COP}\right)}_{\text{Canot}}$ in equation (IV) to get $\text{COP}$.

  $\begin{array}{c}\text{COP}=40.0\%\times \frac{1}{100}\times \left(11.2\right)\\ =4.48\end{array}$

Initial coefficient of performance of the air conditioner was $5.60$ and final is $4.48$.

Use equation (XII) to calculate fractional change.

  $\begin{array}{c}\text{fractional change}=\frac{4.48}{5.60}\\ =0.800\times 100\\ =80.0\%\end{array}$

Thus, final COP is $80.0\%$ of the initial COP. This implies that COP drops by $20.0\%$.

Therefore, fractional change is to drop by $20.0\%$.