# Homework Help: Carnot heat engine - calculate work

1. Mar 25, 2013

### skrat

Using Carnot heat engine we connect two objects with the same mass m, specific heat capacities c. At the beginning the body 1 has temperature $T_{1}$ and body 2 $T_{2}$. Calculate:
a) How much mechanical work can this engine give?

I tried like this, but i believe this is very wrong :(
$dW=-\eta_{c}Q=(1-\frac{T}{T_{1}})mcdT$
$\int_{0}^{W}dW=-\int_{T_{2}}^{T}(1-\frac{T}{T_{1}})mcdT$
$W=(T_{2}+\frac{T_{1}}{2}(T^{2}-\frac{T^{2}_{2}}{2})-T)$

Can anybody please help me to calculate this the right way?

Thanks!!

Last edited: Mar 25, 2013
2. Mar 25, 2013

### TSny

3. Mar 26, 2013

### Andrew Mason

I am not sure you can do it using calculus. But there is an easy way. You just have to recognize that the total change in entropy of the two reservoirs is: ____.

You can determine the final temperature of both reservoirs using this principle. From that you can determine the change in internal energy of the reservoirs. From the first law, you can then determine the amount of work that has been produced.

AM

4. Mar 26, 2013

### skrat

That is actually part b) of my homework: What is the total change of entropy?

I haven' really thought about that part yet, but my idea was to start like this:
$\Delta S_{1}=\int \frac{dQ}{T}=mc\int \frac{dT}{T}$
Where this is change of entropy for one body only, so:
$\Delta S=\Delta S_{1}+\Delta S_{2}=mc(ln\frac{T}{T_{1}}+ln\frac{T}{T_{2}})$ where the first logarithm is negative (assuming that $T_{1}>T_{2}$ and T somewhere in the middle).

And the final temperature is part c)

So if I find out how to calculate the final temperature I can than calculate the change in internal energy and there by from first law also the amount of work produced.
Well I am assuming that the total change of entropy is equal to 0, but I don't really understand why would this be true and even if it is, how can I than determine the final temperature from the equation above?

Thanks!

5. Mar 26, 2013

### Andrew Mason

Correct. So $\Delta S/mc = ln\frac{T}{T_{1}}+ln\frac{T}{T_{2}} = 0$ which reduces to:

$$\ln{\frac{T}{T_{1}}} = \ln{\frac{T_{2}}{T}}$$

So what is T in terms of T1 and T2 (hint: apply the inverse of ln to both sides)?

ΔS = 0 because all cycles are reversible Carnot cycles. The system and reservoirs are in equilibrium at all times.

AM

Last edited: Mar 26, 2013
6. Mar 27, 2013

### skrat

So every reversible cycle has ΔS=0. But Carnot cycle is reversible, thereby the change of total entropy is equal to zero. Right?

Ok, I hope this is right:
$mcln(\frac{T}{T_{1}})+mcln(\frac{T}{T_{2}})=0$
$ln(\frac{T^{2}}{T_{2}T_{1}})=0$
$\frac{T^{2}}{T_{2}T_{1}}=1$
$T^{2}=T_{2}T_{1}$
$T=\sqrt{T_{2}T_{1}}$(result for part c) final temperature)
Where T is always positive.

We already said that ΔS=0, but let's make sure (mainly because it's part b) of my homework):
$\Delta S=mcln(\frac{T^{2}}{T_{1}T_{2}})=mcln(\frac{T_{1}T_{2}}{T_{1}T_{2}})=0$

Let's move to part a) so total amount of work given... hmm weird, but my result is 0?
If I understand correctly, you use letter U for internal energy? Funny. =)
$\Delta U=\Delta Q+\Delta W$
$\Delta W=\Delta U-\Delta Q$
$\Delta W=(mc(T-T_{1})+mc(T-T_{2}))-(mc(T-T_{1})+mc(T-T_{2}))$
$\Delta W=mc(2T-T_{1}-T_{2})-mc(2T-T_{1}-T_{2})$
$\Delta W=0$

Now I will be happy if anybody corrects me again.=)

7. Mar 27, 2013

### TSny

Think about what ΔU should be for the engine as it goes through one cycle. U is a state variable, and when something goes through a cycle it returns to its initial state.

8. Mar 27, 2013

### skrat

Of course! Hah, completely forgot about that.

So $\Delta W=\Delta Q=mc(2T-T_{1}-T_{2})$

9. Mar 27, 2013

### TSny

That's close. Check the overall sign.

10. Mar 27, 2013

### Andrew Mason

That is one way to look at it.
Or you could apply the first law to the whole system of engine + reservoirs as an isolated system, in which ΔQ = 0 (no heat flow into or out of the system). In that case, ΔU = W. The change in internal energy of the reservoirs is just:

ΔU = mC(ΔTh + ΔTc)

AM

11. Mar 28, 2013

### skrat

What if the heat engine works n times worse than carnot engine?

Is in that case total change of entropy still 0 and final temperature the same as before, so: $T=\sqrt{T_{1}T_{2}}$ ?

and $\Delta W=-\eta \Delta Q$ where $\eta =\frac{\eta _{c}}{n}$ ?

12. Mar 28, 2013

### Andrew Mason

I am not sure what n times worse than the Carnot means. The entropy would definitely not be 0 if it is an irreversible process.

AM

13. Mar 29, 2013

### skrat

It means that we have an carnot heat engine that has effciency defined as $\eta _{c}=1-\frac{T_{C}}{T_{H}}$. A similar heat engine that works n times worse that carnot heat engine is an engine whose efficiency $\eta _{x}$ is defined as $\eta _{x}=\frac{1-\frac{T_{C}}{T_{H}}}{n}$.
So Carnot heat engine has maximum efficiency, now I have to calculate the amount of work, final temperature and total change of entropy if the heat engine has lower efficiency.

At least, that't how I understand it...
I hope my english is not too bad to understand it.

14. Mar 29, 2013

### Andrew Mason

So efficiency = W/|Qh| = (|Qh|-|Qc|)/|Qh| = (1/n)(Th-Tc)/Th. This means that:

That appears to reduce to:

|Qh|/Th = |Qc|/Tc - (n-1)W/Tc which, since Qc > 0 and Qh < 0, means:

Qh/Th + Qc/Tc = (n-1)W/Tc

If the the heat reservoirs are such that there is negligible change in temperature of each reservoir in each cycle, the change in entropy in one cycle is:

ΔS = ΔSh + ΔSc = Qh/Th + Qc/Tc = (n-1)W/Tc

AM