The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 11, Problem 3SP

(a)

To determine

The efficiency of the Carnot engine.

(a)

Expert Solution
Check Mark

Answer to Problem 3SP

The efficiency of the Carnot engine is 5.1%.

Explanation of Solution

Given info: Temperature of hot reservoir is 25°C and temperature of cold reservoir is 5°C.

Write an expression to calculate the efficiency.

e=1TCTH

Here,

e is the efficiency of the Carnot engine

TC is the temperature of cold reservoir

TH is the temperature of hot reservoir

Substitute 5°C for TC and 25°C for TH to find e.

e=1(5+273)K(25+273)K=0.067=(0.067)(100%)=6.7%

Thus, the efficiency of the Carnot engine is 6.7%.

Conclusion:

The efficiency of the Carnot engine is 6.7%.

(b)

To determine

The work provided in each cycle.

(b)

Expert Solution
Check Mark

Answer to Problem 3SP

The work provided in each cycle is 13.4 J.

Explanation of Solution

Given info: Heat released to hot reservoir is 200J.

Write an expression for work provided in each cycle.

W=eQH

Here,

QH is the heat released to hot reservoir.

W is the work done.

Substitute 0.067 for η and 200J for QH to find W.

W=(0.067)(200J)=13.4 J

Thus, work provided in each cycle is 13.4 J.

Conclusion:

The work provided in each cycle is 13.4J.

(c)

To determine

The heat released from cold reservoir.

(c)

Expert Solution
Check Mark

Answer to Problem 3SP

The heat released from cold reservoir is 186.6 J.

Explanation of Solution

Write an expression for heat released from cold reservoir.

QC=QHW

Here,

QC is the heat released from cold reservoir

Substitute 200J for QH and 13.4 J for W to find QC.

QC=200J13.4J=186.6J

Thus, the heat released from cold reservoir is 186.6 J.

Conclusion:

The heat released from cold reservoir is 186.6 J.

(d)

To determine

The coefficient of performance of the heat pump.

(d)

Expert Solution
Check Mark

Answer to Problem 3SP

The coefficient of performance of the heat pump is 14.9.

Explanation of Solution

Write an expression for coefficient of performance of the heat pump.

K=QCW

Here,

K is the coefficient of performance of the heat pump

Substitute 200 J for QC and 13.4 J for W to find K.

K=200J13.4J=14.9

Thus, the coefficient of performance of the heat pump is 14.9.

Conclusion:

The coefficient of performance of the heat pump is 14.9.

(e)

To determine

The possibility of application of the heat pump for home heating.

(e)

Expert Solution
Check Mark

Answer to Problem 3SP

Yes, the heat pump can be used for home heating.

Explanation of Solution

For home heating only moderate energy range is required. That will be sufficient to increase the temperature of home slightly. Here for working of the pump, only moderate range of energy is required.

Since the energy required is moderate, the energy provided by the heat pump will be adequate. The energy release will be less compared to the energy required to run the heat pump.

Conclusion:

Yes, the heat pump can be used for home heating.

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Chapter 11 Solutions

The Physics of Everyday Phenomena

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