# Thermodynamics (Carnot Engines)

peaches1221
1. The amount of heat that an engine rejects to the cold reservoir is 3 times the amount of work that it does. What is the efficiency of the engine?

e = W/ Qh = 1 - Qc/Qh

For this problem, I said e= w/3w. Thus e would equal 1/3 = 0.333. This however is not correct. What did I do wrong?

2. A Carnot engine operates between the temperatures 390 K and 300 K. How much heat is discarded for the engine to produce 1900 J of work?

e = 1 - Tc/Th
e = 1 - 300K/390K = 0.2307
e = W / Qh
thus 0.2307 = 1900J/ Qh
I solved for Qh and i got the answer to be 8233.3 J

This isn't correct either, but I'm not sure what I'm doing wrong.

3. During each cycle a reversible engine absorbs 1550 J of heat from a high-temperature reservoir and performs 1200 J of work. How much heat is exhausted to the low-temperature reservoir during each cycle?

Qh = W + Qc
1550 = 1200 + Qc and then I solved for Qc. This isn't correct either.

Homework Helper
Gold Member
You're rushing through these and making algebra errors. How did you go from e = 1 - Qc/Qh to e = W/3W? What is the relationship between W, Qc, and Qh, according to conservation of energy?

Homework Helper
1. The amount of heat that an engine rejects to the cold reservoir is 3 times the amount of work that it does. What is the efficiency of the engine?

e = W/ Qh = 1 - Qc/Qh

For this problem, I said e= w/3w. Thus e would equal 1/3 = 0.333. This however is not correct. What did I do wrong?
Who said e=W/Qc?

It's given that Qc = 3W => e = 1 - 3W/Qh. Also, e = W/Qh.

Eliminate Qh from these two eqns and see how much is e.

Same for the other eqns.