Carnot Heat Engine, thermodynamics

[theown1]

A carnot engine has an efficiency of 59% and performs 2.5x10⁴ J of work in each cycle. a) how much heat does the engine extract from its heat source in each cycle?
b) suppose the engine exhausts heat at room temp (20^oC) what is the temperature of the heat source?



I know that the efficiency of a heat engine is e=W/Q_h or for a carnot engine e =1-T_c/T_h



3. solution
I managed to find the solution for the first part which was 0.59=W/T_h(W=2.5x10⁴J)
and I just solved for T_h which came out to be 4.2x10⁴J

but I'm not sure how I go about solving for part b, I was thinking that since it said the engine exhausts heat at room temp, so does that mean that T_h-T_c=20? or do I have to convert that temperature to internal energy and solve? I'm not sure I'm pretty confused on what it means, can anyone help

 

[RTW69]

I think what this means is Tc=20 C, efficiency stays the same what is the new Th. See if that works for you.

 

[theown1]

ya that worked, thanks!

 

[Andrew Mason]

I managed to find the solution for the first part which was 0.59=W/T_h(W=2.5x10⁴J)
and I just solved for T_h which came out to be 4.2x10⁴J

Careful. I think you meant Qh not Th.

but I'm not sure how I go about solving for part b, I was thinking that since it said the engine exhausts heat at room temp, so does that mean that T_h-T_c=20? or do I have to convert that temperature to internal energy and solve? I'm not sure I'm pretty confused on what it means, can anyone help

First convert temperatures to Kelvin. What is 20C in K?

Look at your equation for efficiency (as a function of Th and Tc). You have Tc (in Kelvin), and you have the efficiency so you can easily find Th.

AM