What Is the Maximum Work a Heat Engine Can Perform in a Cycle?

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SUMMARY

The maximum work a heat engine can perform in a cycle operating between temperatures of 700 K and 400 K, with a heat input of 2800 J, is 1200 J. This conclusion is derived using the Carnot efficiency formula, where the efficiency (η) is calculated as η = 1 - (T_c/T_h). Substituting the given temperatures yields an efficiency of approximately 0.42857. The maximum work done (W) is then calculated as W = η * Q_h, resulting in W = 1200 J.

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Homework Statement



21. A heat engine operates in a cycle between temperatures 700 K and 400 K. The heat input to the engine during each cycle is 2800 J. What is the maximum possible work done by the engine in each cycle?

(A) 1200 J
(B) 1600 J
(C) 2100 J
(D) 2800 J
(E) 4400 J

I believe that the answer is E. I just don't know how to prove it. If you could show that would be great. I know that



Homework Equations



e = |W|/|Qh|

|Qh| = |W| + |Ql|

e = 1 - |Ql|/|Qh|

The Attempt at a Solution



I'm lost...

e = |W|/|Qh|

|Qh| = |W| + |Ql|

e = 1 - |Ql|/|Qh|

I don't remember how to do this type of problem...
 
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Wait the answer is A... but why?
 
GreenPrint said:
Wait the answer is A... but why?
The maximum possible efficiency is in a reversible (Carnot) cycle. In a reversible cycle, \Delta S = 0, so \Delta S = \Delta S_h + \Delta S_c = -Q_h/T_h + Q_c/T_c = 0. This means that Q_c/Q_h = T_c/T_h

So \eta = W/Q_h = (Q_h-Q_c)/Q_h = 1 - Q_c/Q_h = 1 - T_c/T_h

Plugging in the numbers: \eta = 1 - 400/700 = .42857.

Since W = \eta * Q_h, the maximum Work is 1200 J. (.42857 * 2800).

AM
 

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