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A Carnot heat engine and an ordinary refrigerator with coefficient of performance 2.00operate between reservoirs at 350Kand 250K. The work done by the Carnot heat engine drives the refrigerator. If the heat engine extracts 10.0J of energy from the hot reservoir, how much energy does the refrigerator exhaust to the hot reservoir?

Short Answer

Expert verified

The refrigerator rejects8.57Jthermal energy to the hot reservoir.

Step by step solution

01

Step : 1 Given Information

Temperatures of the hot and cold reservoirs between which Carnot heat engine operates are: TH=350Kand TC=250Krespectively. The amount of heat extracted from the hot reservoir (source) is Q1=10J. Let the amount of heat rejected to the cold reservoir (sink) is Q2and the amount of work done by the Carnot heat engine is Wout. This heat engine drives a refrigerator whose coefficient of performance is β=2. Let the amount of heat extracted from the cold reservoir (source) is role="math" localid="1648621324594" Q'2, and the amount of heat rejected to the hot reservoir (sink) is Q'1.

02

Step : 2 Calculation

Formula used:

Efficiency of a Carnot heat engine, =WoutQ1=Q1-Q2Q1=TH-TCTH;Coefficient of performance of Carnot refrigerator, β=Q'2Win=Q'2Q'1-Q'2

Calculation:

Efficiency of the Carnot heat engine : η=Q1-Q2Q1=TH-TCTH

Q1-Q2Q1=350-250350=100350=27

or, 10-Q210=27Q2=507J

03

Step : 3 Calculation of Q'1

Thus, Wout=Q1-Q2=10-507=207J.While amount of work input to the refrigerator Win=Q'1-Q'2=Wout=207JQ'1=207+Q'2.Using this relation betweenQ'1and Q'2in coefficient of performance, β=Q'2Win=Q'2Q'1-Q'2=2

or, Q'2207+Q'2-Q'2=2Q'2=407J

Therefore, Q'1=207+Q'2=207+407=607=8.57J

Conclusion: The refrigerator rejects 8.57J thermal energy to the hot reservoir.

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