Heat Engines and the Carnot Cycle (1 Viewer)

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[SOLVED] Heat Engines and the Carnot Cycle

1. The problem statement, all variables and given/known data

The operating temperatures for a Carnot engine are Tc and Th= Tc + 55K. The efficiency of the engine is 15%.

How do i find Tc?
 

nicksauce

Science Advisor
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[tex]e = 1 - \frac{Tc}{Th} = 1 - \frac{Tc}{Tc+55}[/tex]

One equation one unknown should not be a problem...
 
Thats all mastering physics gives me. A problem like that is in the book but the efficiency of the engine is 11% instead. They get 450 K for Tc
 

dynamicsolo

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Keep in mind that while efficiencies are often quoted as percentages, they must be used in equations as fractions. Set e = 0.15 and you should be able to solve the Carnot equation for Tc.
 
Do i set e= 0.15 to the equation above? and if i do how do i go about getting the answer?
 

dynamicsolo

Homework Helper
1,649
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Do i set e= 0.15 to the equation above?
That's correct.

and if i do how do i go about getting the answer?
You will have to re-arrange the equation algebraically. Starting from

[tex]0.15 = 1 - \frac{Tc}{Tc+55}[/tex] , the first step would be

[tex]\frac{Tc}{Tc+55} = 1 - 0.15 = 0.85[/tex]
 
thanks for the help
 

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