Heat Engine Problem: Q, Entropy, W, Change in Entropy of Universe

[DanAbnormal]

Homework Statement

A reversible heat engine operates between two reservoirs having temperatures T1 and T2 (T2 > T1). The temperature T1 of the cold reservoir remains constant, whereas the warmer reservoir consists of n moles of a gas at constant volume with specific heat capacity Cv. After the heat engine has operated for period of time \Deltat, the temperature T2 has dropped to T1.

i) How much heat is extracted from the warmer reservoir during \Deltat?

ii) What is the change of entropy of the warmer reservoir during \Deltat?

iii) How much work did the engine produce during \Deltat?

iv) What is the change in entropy of the universe during \Deltat?

Homework Equations

Q = nCv\DeltaT

The Attempt at a Solution

i) Right for the first part I am assuming \DeltaT corresponds to (T2 - T1), so the heat extracted is

Q = nCv(T2 - T1) ?

ii) For the next bit I am not sure. It says the heat engine is reversable, which makes me think there's no change in entropy, however it hasnt completed a cycle in \Deltat. So that can't be right... can it?

iii) For the work done W = Qh - Qc

and Qh (if my attempt at part one is right) is = nCv\DeltaT

so W = (nCv\DeltaT) - Qc ?

iv) And the last part is kind of the same problem I have with part two.

So for the bits I've done am I right, and for the entropy parts is it zero or not?

 

[Topher925]

1. That makes sense to me.

2. The question asks what's the change in entropy of the reservoir not the engine. S = int(Q/dt) + (s2-s1)

3. Uhhh...I'm not to sure, be careful with your deltaT. I will agree with W = Qh-Qc

4. I have no idea how the hell anyone would find this but if they did that would be pretty amazing. I'm just going to guess and say the answer is, The Letter H.

 

[DanAbnormal]

2. The question asks what's the change in entropy of the reservoir not the engine. S = int(Q/dt) + (s2-s1)

what does s2 and s1 mean?