Carnot engine efficiency problem

[Adriane Baun]

Homework Statement

A Carnot engine absorbed 1.0 kJ of heat at 300 K, and exhausted 400 J of heat at the end of the cycle. What is the temperature at the end of the cycle?

Homework Equations

The efficiency of a Carnot engine is given by the formula
Efficiency = 1 – Qc/Qh
= 1 – Tc/Th

The Attempt at a Solution

Efficiency = 1 – 300J/1000J
= 0.7 = 70%
0.7 = 1 – Tc/Th = 1 – Tc/300K
Tc = 90K
Is it correct? I am trying to self study the topic Carnot cycle for a report

 

[TSny]

Welcome to PF!

The Attempt at a Solution

Efficiency = 1 – 300J/1000J
= 0.7 = 70%
0.7 = 1 – Tc/Th = 1 – Tc/300K
Tc = 90K
Is it correct?

Did you mean to use 300 J for Q_C? The problem statement gives 400 J.

Otherwise, I think your calculation is correct. However, to specify the temperature at the end of the cycle, you need to know which point of the cycle is taken to be the "end of the cycle".

 

[Adriane Baun]

Oh i see i misread the problem so it should be
Efficiency = 1 – 400J/1000J = 0.6
Then
0.6 = 1 – Tc/Th = 1 – Tc/300K
Tc = 120K

 

[TSny]

OK. I'm still not clear on which point of the cycle is the endpoint. So I'm not sure if T_C is the temperature at the endpoint. But your work for T_C looks good.