# The Carnot Icemaker heat question

1. Dec 8, 2007

### xinlan

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

An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0C and rejects heat to a room at a temperature of 25.0C. Suppose that liquid water with a mass of 78.3kg at 0.0C is converted to ice at the same temperature.

Take the heat of fusion for water to be = 3.34×105 J/kg.

How much heat |Qh| is rejected to the room?

2. Relevant equations

|Qh| = Mwater*Cwater*.(Tf-Ti) + Mwater*Lfusion

3. The attempt at a solution

I have really no idea to answer this..
I just guessed that equation since the heat of fusion is involved..

please help me.. thank you so much..

2. Dec 8, 2007

### cepheid

Staff Emeritus
The temperature of the water doesn't change. It goes from liquid water at 0.0C to ice at 0.0C. So only a phase change occurs. So the amount of heat you have to take out of the liquid water would just be given by M*Lfusion.

But here's where you have to be careful. The second law of thermodynamics states that you can't transfer heat from a cooler body to a hotter body without an additional input of mechanical work. So, the quantity that we just calculated above, the heat that must be taken from the cool reservoir (icemaker) is |Qc|. However, work must be done in order to do so, so that the heat rejected is greater than |Qc|:

|Qh| = |Qc| + |W|

You don't know how much work is expended doing so, but IT CAN BE SHOWN THAT for a carnot cycle, this depends solely on the temperature difference between the cool and hot reservoirs. In other words, you can calculate the coefficient of performance (K) of your icemaker in terms of Th = 25C and Tc = 0C. You can look up how K depends on Tc and Th in your notes. This coefficient of performance will be:

K = |Qc| / |W| = |Qc| / (|Qh| - |Qc|)

and from it, it's clear that you can deduce how much greater the heat rejected is than the heat extracted.

Does this make sense? Do you understand why the efficiency of the Carnot cycle comes into this problem?

Last edited: Dec 8, 2007
3. Dec 9, 2007

### xinlan

I calculated the amount of heat taken out of the liquid water
Q = M*L
is that Q is Qc?

then..
the coefficient of performance can be gotten from (Th-Tc)/Th ?
isn't K = W/Qh ? so, Qh = W/K

I am confused..

4. Dec 9, 2007

### cepheid

Staff Emeritus
Well, if you got these formulas from your notes or text then they must be correct.

When I looked it up, I found that K = |Qc|/|W|...maybe your book defines it differently. That is sort of irrelevant...use whatever def'n of K you have been given. Either way, we can solve for Qh

The point is, you know the temperatures, so you can solve for K. But you DON'T KNOW |W|, so it makes sense that you should use a form in which |W| is eliminated:

by definition, |W| = |Qh| - |Qc| -- it is the amount of extra energy that must be put in to make this process happen. Substituding this into (my version of ) K, we get:

K = |Qc| / (|Qh| - |Qc|)

Can you see that |Qh| is now the only unknown? This means that we can solve for it by rearranging the equation using simple algebraic manipulations in order to isolate |Qh| in terms of the two knowns.

EDIT: Okay...I've cleared up the confusion. The two versions for the coeff. of performance are chosen depending on your point of view i.e. is the device being used as a fridge or a heat pump? For a fridge, my formula for K is correct: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatpump.html#c3

Let me know if that clears things up, or if you need more help

Last edited: Dec 9, 2007
5. Dec 9, 2007

### xinlan

I understood now..
my teacher hasn't taught me the coefficient of performance of the fridge. that's why I was confused it with efficient of heat engine..

anyway.. I got it now..
thank you so much for helping me..

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook