Reversible(Carnot) Cycle Temperature Equilibrium Problem

[kalbuskj31]

Homework Statement

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.

Homework Equations

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

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

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.

 

[Mindscrape]

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

 

[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

 

[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.

 

[paranormal]

Hello,

I would like to clarify a few things about the problem and your solution attempt. First, let's define some terms and concepts.

A reversible Carnot cycle is a theoretical thermodynamic cycle that operates between two heat reservoirs at different temperatures. In this cycle, the system goes through a series of reversible processes (i.e. processes that can be reversed without any loss of energy) to convert heat into work. This cycle is often used as a benchmark for the maximum efficiency that any heat engine can achieve.

Now, let's look at the problem. We have a reversible heat engine that operates in a cycle, withdrawing heat from a high-temperature reservoir (T1) and rejecting heat into a low-temperature reservoir (T2). The heat engine performs work (W) and the two reservoirs have constant heat capacities (C1 and C2).

To calculate the final temperature of the system (Tf) and the maximum work done by the engine (Wmax), we can use the following equations:

W = Q2 - Q1 (first law of thermodynamics)
Wmax = Qrev (maximum work done in a reversible process)
ΔU = ΔH = 0 (for a reversible process)
C = Q/ΔT or dQ/dT (heat capacity)
Q = nCpΔT (heat transfer)

Now, let's look at your solution attempt. You correctly calculated the heat transferred from each reservoir (q1 and q2) and the work done by the engine (w). However, you made a mistake in the sign of the C2 term in your calculation of w. The heat from the high-temperature reservoir is being added to the heat engine, so it should be positive. Thus, the correct equation for w is:

w = C2(Tf-T2) - C1(Tf-T1)

Next, you correctly identified that the process continues until T1 = T2 = Tf. This means that the temperature change (ΔT) in both reservoirs is the same, and we can simply write it as ΔT. Now, we can use the equations for heat capacity and heat transfer to relate the temperatures and heat transfers:

q2 = nC2ΔT
q1 = nC1ΔT

Substituting these into the equation for work, we get:

w = nΔT(C2 - C1)(Tf -