Heat Engines and the Carnot Cycle

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Homework Help Overview

The discussion revolves around a problem involving a Carnot engine, specifically focusing on determining the cold reservoir temperature (Tc) given the efficiency and the relationship between the hot and cold temperatures.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between efficiency and temperature in the context of the Carnot cycle, with some questioning how to correctly apply the efficiency value in calculations. There are also considerations about the differences in efficiency values provided in various sources.

Discussion Status

Participants are actively engaging with the problem, with some offering guidance on how to set up the equations. There is a recognition of the need to rearrange the equation algebraically, and while some questions remain about the process, a productive direction has been established.

Contextual Notes

There is mention of discrepancies in efficiency values from different sources, which may affect the interpretation of the problem. Additionally, the need to convert efficiency percentages into fractions for calculations is highlighted.

HELLO11
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[SOLVED] Heat Engines and the Carnot Cycle

Homework Statement



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?
 
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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
 
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?
 
HELLO11 said:
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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