Heat Engines and the Carnot Cycle

[HELLO11]

[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?

 

[nicksauce]

$$e = 1 - \frac{Tc}{Th} = 1 - \frac{Tc}{Tc+55}$$

One equation one unknown should not be a problem...

 

[HELLO11]

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]

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.

 

[HELLO11]

Do i set e= 0.15 to the equation above? and if i do how do i go about getting the answer?

 

[dynamicsolo]

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

$$0.15 = 1 - \frac{Tc}{Tc+55}$$ , the first step would be

$$\frac{Tc}{Tc+55} = 1 - 0.15 = 0.85$$

 

[HELLO11]

thanks for the help