EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100454897
Author: Jewett
Publisher: YUZU
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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.

(a)

Expert Solution
Check Mark
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 8.48kW .

Explanation of Solution

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

The formula for Carnot efficiency of cooling is,

Ec=TlToTl

Here,

Ec is Carnot efficiency of cooling.

Tl is room temperature.

To is temperature outside the room.

Substitute 7.00°C for Tl and 27.0°C for To in the above expression.

Ec=7.00°C27.0°C7.00°C

Solve the above expression for Ec ,

Ec=(7.00°+273°)K(27.0°+273°)C(7.00°+273°)K=280.0K300.0K280.0K=14

The coefficient of performance at 27°C  is equals to 40% of Ec .

COP27°C=40%(Ec) ,

Substitute 14 for Ec in the above expression.

COP27°C=40%(14)=5.6

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

Qin+W=Qout

Rearrange the above expression for Qin ,

Qin=QoutW (1)

The formula to coefficient of performance is,

QinW=C.O.P27°C

Here,

C.O.P27°C is coefficient of performance.

Qin is the rate of removal of energy.

W is input work.

Substitute QoutW for Qin in the above expression.

QoutWW=C.O.P27°C

Solve the above expression for W ,

QoutW1=C.O.P27°C

Substitute 10.0kW for Qout and 5.6 for C.O.P27°C in the above expression.

10W1=5.6W=1.515KW (2)

Substitute 1.515kW for W and 10.0kW for Qout in equation (1).

Qin=(10.0KW)(1.515KW)=8.48kW

Conclusion:

Therefore, the rate of removal of energy by the air conditioner is 8.48kW .

(b)

Expert Solution
Check Mark
To determine

The power required for the input work.

Answer to Problem 22.70AP

The power required for the input work is 1.52kW .

Explanation of Solution

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

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

W=1.515kW1.52kw

So the work input is 1.52kW .

Conclusion:

Therefore, the work input required for the input work is 1.52kW .

(c)

Expert Solution
Check Mark
To determine

The entropy change of universe produced by air conditioner in 1.00h .

Answer to Problem 22.70AP

The entropy change of universe produced by the air conditioner in 1.00h is 1.09×104J/K .

Explanation of Solution

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

The formula to calculate change in entropy is,

ΔS=(QoutToQinTl)t

Substitute 10.0kW for Qout , 300°K for To , 8.485kW for Qin , 280°K for Tl and 1.00h for t in the above expression.

ΔS=(10.0kW300°K8.485kW280°K)(1.00h)

Solve the above expression for ΔS .

ΔS=(10.0kW×(103W1kW)300°K8.485kW×(103W1kW)280°K)(1.00h×(3600s1.00h))=10907.14=1.09×104J/K

Conclusion:

Therefore, the entropy change of universe produced by the air conditioner in 1.00h is 1.09×104J/K .

(d)

Expert Solution
Check Mark
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°C . The temperature outside the laboratory is 27.0°C . The rate at which air conditioner emits energy outside is 10.0kW . The C.O.P of model is 40.0% . The time for the air conditioner is 1.00h . The temperature outside increases to 32.0°C .

The formula to calculate COP is,

COP32°C=TlToTl

Here,

COP32°C is coefficient of performance at 32°C .

Substitute 7.00°C for Tl and 32.0°C for To in the above expression.

COP32°C=7.00°C32.0°C7.00°C=(273°+7.0°)K(273°+32.0°)K(273°+7.0°)K=280°K305°K280°K=11.2

The formula for the percentage is,

COPchange=(COP27°CCOP32°CCOP27°C)(100)

Here,

COPchange is the change in coefficient of performance as compared to the previous coefficient of performance.

COP32°C is coefficient of performance at 32°C .

Substitute 14 for COP27°C and 11.2 for COP32°C in the above expression.

COPchange=(1411.214)(100)=20%

Conclusion:

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

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Chapter 22 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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