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

AI Thread Summary
A heat engine operating between temperatures of 700 K and 400 K with a heat input of 2800 J can achieve a maximum work output based on its efficiency. The maximum efficiency for a reversible (Carnot) cycle is calculated as η = 1 - (T_c/T_h), resulting in an efficiency of approximately 0.42857. By applying this efficiency to the heat input, the maximum work done by the engine is determined to be 1200 J. The confusion in the discussion stems from miscalculating the answer, initially suggesting 4400 J before correcting to 1200 J. The final conclusion is that the maximum possible work done by the engine in each cycle is 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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