# Homework Help: Entropy and Carnot refrigerators

1. Apr 13, 2014

### CAF123

1. The problem statement, all variables and given/known data
a)A stone at 400K with heat capacity $c_p$ is placed in a very large lake at 300K. The stone cools rapidly to 300K. Calculate the entropy change, due to this process, of the stone and lake.

b)An insulated cool-box of a Carnot refrigerator at temperature T loses heat at a rate proportional to the difference between T and the atmosphere To, i.e $\dot{Q} = B(T_o-T)$ where B is some constant. Show that the temperature of the cool-box is given by the solution of $T^2 - \left(2T_o + \frac{W}{B}\right)T + T_o^2 = 0$

2. Relevant equations
Definition of entropy $dS = dQ/T$. Carnot refrigerator efficiency

3. The attempt at a solution
a)The stone may be regarded as the system and the lake as the surroundings. The lake is effectively a reservoir and so the addition of the stone to the lake will alter the temperature of the lake negligibly. When the stone is placed in the lake, there will be a net flow from lake → stone until they come to thermal equilibrium whereby the heat flow is equilibrated.
$dS_{stone} = \frac{dQ_{res}}{T} = \frac{c_p dT}{T}$. $dQ_{res}$ is the amount of heat the reservoir (or lake) provides to the stone to change its temperature to 300K. Evaluating gives $\Delta S = c_p \ln(3/4) < 0$ as expected. Intuitively, the entropy of the reservoir should change negligibly, but I don't think I am supposed to ignore such a change.

By the second law, $\Delta S_{stone} + \Delta S_{lake} \geq 0$ and in this case, no equality since the process is irreversible. So $\Delta S_{lake}$ must be positive to accommodate this. (albeit very small - the addition of the rock increases the entropy of the lake not significantly). How would I calculate this?

b)My starting point was to use the Carnot efficiency $Q_1/(Q_1 - Q_2) = T_1/(T_1-T_2)$, where $Q_1$ is the heat taken from the refrigerator and $Q_1-Q_2$ is the work provided to run the refrigerator. But I am not sure how to use this and/or the expression given.
Thanks.

2. Apr 13, 2014

### dauto

Your intuition is mistaken. The lake's temperature doesn't change so the entropy calculation for the lake is just Slake = Q/Tlake.

3. Apr 13, 2014

### rude man

In part (b), what is W?

4. Apr 13, 2014

### CAF123

Q gained by stone = Q lost by lake. So $Q = c_p (400-300) = 100c_p$. Which means $\Delta S_{lake} = c_p/3$ and so indeed the sum $\Delta S_{stone} + \Delta S_{lake} > 0$. Is that correct? Because of the size of the lake, this change in entropy of the lake is negligible $(c_{p,stone} << c_{p,lake}$).
W was given in the next part, sorry, it is the power input. So, in a Carnot refrigerator cycle, it represents the work done on the refrigerator to enable it to extract heat from the interior and dump it to the outside world (i.e somewhere at a higher temperature, or reservoir.)

Last edited: Apr 13, 2014
5. Apr 13, 2014

### dauto

Yes, that seems correct. The only thing bothering me is that the mass of the stone doesn't show up anywhere in that problem, and it should. Instead of cp we should've been using mcp all along.

6. Apr 14, 2014

### CAF123

I think I interpreted W correctly, the question doesn't really say only that it is a power input. I suppose another interpretation is $W=\dot{Q}$, but I think the question would have made that clearer.

7. Apr 14, 2014

### rude man

W is usually work done by the system. So in a Carnot refrigeration cycle W is the net work done by the coolant in one cycle.

I and I'm sure others will pitch in to help you further if necessary - stand by!

8. Apr 14, 2014

### CAF123

See sketch. So W is the work done by the coolant in extracting heat from the interior of the refrigerator per cycle.

From the efficiency equation, this corresponds to $Q_1 - Q_2 = W$. So my equation is $$\frac{Q_1}{W} = \frac{T_o}{T_o-T}$$ In this case, $Q_1$ corresponds to the heat transferred to the outside world, which would correspond to the integral of $\dot{Q}$ given in the question. So $Q = B(T_o - T)t$. Subbing this in to my efficiency formula gives $$(T^2 + T_o^2 - 2T_oT)t - T_o \frac{W}{B} = 0,$$ but that is not quite the result.

#### Attached Files:

• ###### Refrigerator.png
File size:
1 KB
Views:
90
9. Apr 14, 2014

### rude man

You're almost there.
Efficiency for a refrigerator is not what you stated. Think again about it. What are you trying to do per cycle, and what is the cost of doing it?

10. Apr 14, 2014

### CAF123

A general way to express efficiency would be to say it is (what you want out/what you put in). In this case, we want heat out of the refrigerator and, for this to happen, the coolant must do work W per cycle. Therefore efficiency = Q_2/W = (Q-W)/W, where Q_2, Q_1=Q are defined in the sketch.

But this gives me $$(T^2 + T_o^2 - 2T_oT)t - 2T_o W/B + TW/B = 0$$

11. Apr 14, 2014

### rude man

OK, I should have noticed earlier that here W is not the work done per cycle (as conventional) but the work done in unit time.

So instead of Q2 = Q1 - W it's
Q2dot = Q1dot - W
and there's then no "t" in your expression.

You already realize that you have to equate heat removed in unit time to heat entered in in unit time and I think you'll get the right answer this time. (I can confirm the answer.)

12. Apr 14, 2014

### CAF123

Since we are taking a ratio, the result $\dot{Q_2}/W = T_o/(T_o-T)$ holds. This is equal to $(\dot{Q_1} - W)/W = T_o/(T_o-T)$. But this will give me the same result before, just without any t dependence. I seem to just have an extra term when I simplify it.

13. Apr 14, 2014

### rude man

Sorry to disappoint, but that equation is incorrect. Close, but no panatela.

14. Apr 15, 2014

### CAF123

Ok, I found the error and now have the result. What is the physical significance of the two solutions for T? Let W = 1200 J/s and T_o =300K. Then you get two temperatures T=785, 114K.

The solution rejects 785K on the basis that this simply cannot be a refrigerator temperature. I am not sure I like this reasoning (albeit being true!). We have an equation for T and we solve it. If T turned out to be complex or negative then since we are solving a mathematical equation for a physical situation, sure it makes sense to reject such a solution. Is it okay to just say that physically we have a refrigerator and so the only sensible temperature would be the 114K?

Last edited: Apr 15, 2014
15. Apr 15, 2014

### rude man

You have defined Q1 = heat absorbed in reservoir 1 at temperature To, and Q2 = heat removed from reservoir 2 at temperature T, and combine that with the 1st and 2nd laws, i.e.
Q1 = Q2 + W and
Q1/T1 > Q2/T2

means To > T must obtain or W < 0 which violates Clausius's statement of the 2nd law.

16. Apr 15, 2014

### CAF123

Yes, that would be a more satisfying argument. Although I am not quite sure how you obtained the contradiction.
By Clausius, in the notation in the question, $$\frac{Q_1}{T_o} = \frac{Q}{T_o} + \frac{W}{T_o} > \frac{Q}{T} \Rightarrow \frac{Q(T_o-T)}{T} < W$$ What did you do from here?

By also looking at the efficiency formula T/(T0-T), if To < T then the efficiency is negative, so that cannot be the case.

17. Apr 15, 2014

### rude man

Let "1" refer to the hot reservoir (the ambient) and "2" to the cold box interior;

consider the refrigerator: per any integer number of cycles, heat Q2 is removed from the box at temperature T, work W is done on the system, and heat Q1 is pushed to the outside at temperature To.

Assuming all quantities positive, the 1st law says Q1 = Q2 + W.

But: entropy change of the universe must be positive, or Q1/To > Q2/T
and with the 1st law gives (Q2+W)/To > Q2/T
Q2/To + W/To > Q2/T
W > Q2(To/T - 1)

so if T > To then W would be negative which is impossible since we specified at the beginning that W is positive, i.e. the 1st law would be violated..

18. Apr 15, 2014

### CAF123

Yes, that is the expression I obtained above. But the condition T > To (i.e To/T < 1) only implies that To/T - 1 < 0 and so putting it together means that the statement is saying that W is bigger than some negative number. I don't see how it necessarily implies that W is negative. Thanks.

19. Apr 15, 2014

### rude man

I agree, this is badly put.

But the 2nd law permits replacing the > sign with an = sign, as in your Carnot cycle. Then W would be negative.

20. Apr 15, 2014

### rude man

Another view: if T > To then heat would flow from the box to ambient all by itself. If any work were involved, that would by the 1st law have to be positive output work, but we have defined W to be input work rather than output so W would have to be negative, but that would contradict Clausius.

21. Apr 15, 2014

### rude man

OR:

Assuming T > To: (Pretend all Q are Q-dot)
For a Carnot or other fully reversible cycle, sum of entropies into the universe = 0.
Therefore, entropy into the ambient = [B(T-To) + Q1]/To = entropy removed from the box = [B(T-To) + Q2]/T
& note that B > 0.

But Q1 = Q2 + W (W is work ON the system, Q1 flows into the ambient and Q2 flows out of the box).

So [B(T-To) + Q2 + W]/To = [B(T-To) + Q2]/T

and from this you can see that since T > T0, W cannot be positive.