We will consider two applications of the Carnot Cycle of an ideal gas. The gener

Question

We will consider two applications of the Carnot Cycle of an ideal gas. The general diagrams for an engine and a refrigerator are shown here:

First consider a Carnot engine that operates between a hot reservoir at 49°C and a cold reservoir at -15°C. In running 22 minutes, this engine absorbs 540 J of heat from the hot reservoir.

1)

How much heat does it expel to the cold reservoir?

Qc =

2)

How much work does it do in this amount of time?

Wby =

3) What is the upper limit on the efficiency of any heat engine operating between these two reservoirs?

max =

4)

This same system is now run in reverse as a refrigerator between the same two reservoirs. How much heat does it extract from the cold reservoir when 103 J of work are done on the system?

Qc =

J  

5)

How much heat is expelled to the hot reservoir in this process?

Qh =

J

6)

With a refrigerator you pay for the work, and what you get is heat removal. The 'coefficient of performance' of a refrigerator is then defined to be the heat removed from the cold region divided by the work used. What is the coefficient of performance of this refrigerator?

K =

Explanation / Answer

1)

Here ,

Th = 49 degree C = 322 K

Tc = -15 degree C = 258 k

Qh = 540 J

1) as for a carnot cycle ,

1 - Qc/Qh = 1 - Tc/Th

1 - Qc/540 = 1 - 258/322

solving for Qc

Qc = 432.7 J

the expelled to the cold reservior is 432.7 J

2)

Wby = Qh - Qc

Wby = 540 - 432.7

Wby = 107.3 J

the work done is 107.3 J

3)

for the maximum efficiency = 1 - Tc/Th

maximum efficiency = 1 - 258/322

maximum efficiency = 0.198

4)

W = 103 J

for Qc

Qh/W = Th/(Th - Tc)

Qh/103 = 322/(322 - 258)

Qh = 518.22

for Qc = 518.22 - 103

Qc = 415.22 J

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