# Heat Engines and the Carnot Cycle (1 Viewer)

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#### HELLO11

[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

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

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

Homework Helper
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

Homework Helper
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

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