# Homework Help: Reversible(Carnot) Cycle Temperature Equilibrium Problem

1. Sep 10, 2010

### kalbuskj31

1. The problem statement, all variables and given/known data

A reversible hear engine, operating in a cycle, withdraws heat from a high temperature reservoir (temperature consequently decreases). It performs work w, and rejects heat into a low-temperature reservoir(temperature consequently increases).

The 2 reservoirs are initially at temperatures T1 and T2 and have constant heat capacities C1 and C2. Calculate the final temperature of the system and the Wmax from the engine.

2. Relevant equations

W = q2 - q1
Wmax = Qrev
ΔU = ΔH = 0
C = q/ΔT or dq/DT
q = nCpDT

(q2 - q1)/q2 = (t2 - t1)/t2

3. The attempt at a solution

I'm having a hard time understanding what I'm doing from a conceptual perspective and in terms of explanation.

q2 = 1 mole * C2 * (Tf-T2)
q1 = 1 mole * C1 * (Tf-T1)

w = q2 - q1
w = C2(Tf-T2) - [C1(Tf-T1)] **(The answer has the C2 term negative, Is it because heat from the reservoir is added to the heat engine?)

This process continues until T1=T2 @ T.

Efficiency = w/q2 = 1 - q1/q2.

dW/dq_2 = (q2 - q1) / q2 = (T2 - T1) / T2

dW = (T2 - T1) / dQ_2

**I get stuck here. Do we split this into 2 integrals (1 for dT1 and the other for dT2?) I have to find a way to relate the heat q2 to the temperatures of T1 and T2. (q2 = T - T2 and q2T1 = q1T2, am I on the right track?). Help would be much appreciated.

2. Sep 13, 2010

### Mindscrape

Work is just efficiency multiplied by qheat, by definition. You've even written the equation out already, why take it to differentials?

3. Sep 14, 2010

### Andrew Mason

This is a non-trivial problem if you want an exact answer.

You want to find the final common temperature of the two reservoirs. For an initial heat flow of dQh out of the hot reservoir, there is a corresponding temperature decrease of dTh = dQh/C1. There is a heat flow into the cold reservoir of dQc = dQh(Tc/Th) and a temperature increase of dTc = dQc/C2 = dQh(Tc/Th*C2).

The problem is that the heat flow, and consequential temperature change, changes with Tc/Th. That is quite a task to compute.

One could get a reasonable estimate by averaging the initial and final efficiencies (1-Tc/Th and 0) to get an average efficiency of (1-Tc/Th)/2.

Furthermore, it would be difficult to create a reversible cycle here. Heat would have to flow into the engine at a temperature that decreases with temperature of the hot reservoir, rather than isothermally. Similarly, it would have to flow out at a temperature that increases with the temperature of the cold reservoir. In doing so, you lose reversibility. So it is implicit in the question that the reservoirs have sufficient heat capacity that the heat flow in one cycle creates a very small change in temperature of the reservoirs.

AM

Last edited: Sep 14, 2010
4. Sep 15, 2010

### kalbuskj31

Thanks for the help. It took me awhile to understand everything, but I was able to get the correct answers. The material is so dense that sometimes you forget some of the simplest steps.