# Problem with refrigerator and radiator

## Homework Statement

The following data refer to an electrically operated refrigerator:
- Efficiency : ## \xi = 2.4##
- Temperature inside: ##T_i = -9 ° C ##
- Temperature of the radiator ## T_r = 40 °C ##
- Room temperature: ## T_s = 35 °C##
- Total surface of walls: ## A = 3.2 m ^ 2 ##
- Average thickness of the walls: ## d = 4.0 cm##
- Thermal conductivity of the walls:## k = 2.0 · 10^-5 kcal / (m K s) ##
determine:
a) Required power to make this refrigerator work;
b) The change in entropy of the universe in one second, assuming that the room and the radiator are separated (no heat exchange between them);
c) The change in entropy of the universe in one second, assuming that the radiator exchanges heat with the room, remaining a constant temperature (the room is a thermal reservoir).

Answers##[(a) P = 122.8 W; (b) ΔS_u = 0.377 J / K; (c) ΔS_u = 0.399 J / K] ## [/ quote]

## Homework Equations

##\xi=\frac{Q_{absorbed}}{|W_{absorbed}|}##

## The Attempt at a Solution

What confuses me the most is the presence of the radiator: how can I deal with it?
Should I treat it as a further source of heat besides the room or somehow else?
Furthermore I don't understand what is the correct way to get a solution to the first question: I know that the machine absorb a power ##P'=\frac{P}{\xi}##, should I impose that ##P'## equals the flux of heat between room-fridge and radiator-fridge or something else?

A suggestion on how to set up a solution would be highly appreciated!

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

haruspex
Homework Helper
Gold Member
A suggestion on how to set up a solution would be highly appreciated
You can calculate the rate at which heat leaks in. That gives you the rate at which it must be pumped out. You know the temperature gradient against which it must be pumped, so you can calculate the power needed if it were 100% efficient.

Soren4
You can calculate the rate at which heat leaks in. That gives you the rate at which it must be pumped out. You know the temperature gradient against which it must be pumped, so you can calculate the power needed if it were 100% efficient.

Thanks a lot for the reply to my question!!

I thought about the problem, and, as you suggested, I calculated the power needed with success. I also reached the solution to point b), but I'm still struggling on point c).

In the following I will indicate as "refrigerated space" what is called "inside" in the question.

As I said in the question, the problem is that I'm not sure about the role of the radiatior. I interpreted it in this way. The heat is taken from the refrigerated space from the refrigerator and it is given to the radiator, which may or may be not exchange heat with the room. I summed it up in this drawing, indicating with arrows the heats and work.

For point b) the grey arrow in the drawing is not present so I calculated the change in entropy (in one second) in this way

##\Delta S_{refrigerated \, space} = 0 ##

##\Delta S_{radiator}= \frac{k A (T_{room}-T_{refrigerated \, space})}{d (40+273.15)} (1+\frac{1}{2.4})##

##\Delta S_{room}= -\frac{k A (T_{room}-T_{refrigerated \, space})}{d (35+273.15)} ##

##\Delta S_{universe}= \Delta S_{radiator}+\Delta S_{room}=0.377 J/K##

For point c) (grey arrow present) I tried in this way

##\Delta S_{refrigerated \, space} = 0 ##

##\Delta S_{radiator}= \frac{k A (T_{room}-T_{refrigerated \, space})}{d(40+273.15)} (1+\frac{1}{2.4})- \frac{k A (T_{refrigerator}-T_{room})}{d(40+273.15)} ##

##\Delta S_{room}= -\frac{k A (T_{room}-T_{refrigerated \, space})}{d(35+273.15)} + \frac{k A (T_{refrigerator}-T_{room})}{d(35+273.15)}##

##\Delta S_{universe}= \Delta S_{radiator}+\Delta S_{room}=0.378 J/K##

But these last equations do not give the correct result. In this last point I assumed that the exchange of heat between radiator and room happens because of their different temperatures, and I do not know if that is right.

Can you help me in finding the mistake in the resolution of point c) ?

haruspex
Homework Helper
Gold Member
Thanks a lot for the reply to my question!!

I thought about the problem, and, as you suggested, I calculated the power needed with success. I also reached the solution to point b), but I'm still struggling on point c).

In the following I will indicate as "refrigerated space" what is called "inside" in the question.

As I said in the question, the problem is that I'm not sure about the role of the radiatior. I interpreted it in this way. The heat is taken from the refrigerated space from the refrigerator and it is given to the radiator, which may or may be not exchange heat with the room. I summed it up in this drawing, indicating with arrows the heats and work.
View attachment 102510
For point b) the grey arrow in the drawing is not present so I calculated the change in entropy (in one second) in this way

##\Delta S_{refrigerated \, space} = 0 ##

##\Delta S_{radiator}= \frac{k A (T_{room}-T_{refrigerated \, space})}{d (40+273.15)} (1+\frac{1}{2.4})##

##\Delta S_{room}= -\frac{k A (T_{room}-T_{refrigerated \, space})}{d (35+273.15)} ##

##\Delta S_{universe}= \Delta S_{radiator}+\Delta S_{room}=0.377 J/K##

For point c) (grey arrow present) I tried in this way

##\Delta S_{refrigerated \, space} = 0 ##

##\Delta S_{radiator}= \frac{k A (T_{room}-T_{refrigerated \, space})}{d(40+273.15)} (1+\frac{1}{2.4})- \frac{k A (T_{refrigerator}-T_{room})}{d(40+273.15)} ##

##\Delta S_{room}= -\frac{k A (T_{room}-T_{refrigerated \, space})}{d(35+273.15)} + \frac{k A (T_{refrigerator}-T_{room})}{d(35+273.15)}##

##\Delta S_{universe}= \Delta S_{radiator}+\Delta S_{room}=0.378 J/K##

But these last equations do not give the correct result. In this last point I assumed that the exchange of heat between radiator and room happens because of their different temperatures, and I do not know if that is right.

Can you help me in finding the mistake in the resolution of point c) ?
I tried what I think should be a simpler way to get from b to c: what entropy increase is there as the heat delivered to the radiator leaks back into the room?
You may be able to correct some of my numbers. I calculated the heat moved from the room, through the refrigerator and back to the room as 33.6J. Plus the work input to the refrigerator is the 122.8J you found in part a gives 156.4J. As this leaks back from radiator at 313K to room at 308K the net entropy shoild increase by .008 J/K. But according to the answer you quote, that needs to be .022.

Soren4