Carnot Refrigerator Homework: Coeff of Performance, Work & Heat

In summary, the question is asking for the coefficient of performance (COP) of a Carnot refrigerator operating between -30.0 ^ C and + 20.0 ^ C, which is the ratio of the heat removed from the cold reservoir to the work added to the system. The problem provides the heat extracted from the cold reservoir (300 J/s) but does not give the work added to the system. To solve this, you can use the fact that a Carnot refrigerator follows the equation Q1/Q2 = T1/T2, where Q1 is the heat removed from the cold reservoir and Q2 is the heat exhausted to the hot side. By using a block diagram, you can find the relation between Q1,
  • #1
dragon162
17
0

Homework Statement


A Carnot refrigerator operating between -30.0 ^ C and + 20.0 ^ C extracts heat from the cold reservoir at the rate 300 J/s. What are (a) the coefficient of performance of this refrigerator, (b) the rate at which work is done on the refrigerator and (c) the rate at which heat is exhausted to the hot side?


Homework Equations





The Attempt at a Solution


As you can tell I did not include the relevant equations due to the fact that I am very confused here. All I know is that for part A the coefficent of performance of a refrigerator is the ratio of the heat removed from the cold reservoir to the work added to the system. The problem tells me the heat extracted from the cold reservoir but I don't know how to get the work added tot he system. Any help would be appreciated.
 
Physics news on Phys.org
  • #2
Use the fact it's Carnot refrigerator (as opposed to a non-ideal refrigerator). You have enough given information to calculate the coefficient of performance for such a refrigerator.
 
  • #3
Yeah, you know that e=Qout/Win. In the case of a carnot cycle Q1/Q2 = T1/T2. You can use a block diagram to get the Q1, Q2, and work relation.
 
  • #4
Mindscrape said:
Yeah, you know that e=Qout/Win. In the case of a carnot cycle Q1/Q2 = T1/T2. You can use a block diagram to get the Q1, Q2, and work relation.
The COP for a heat pump and a refrigerator are different. If Qout is the heat delivered to the hot register (the outside), you are using the COP for a heat pump. For a refrigerator,

COP = Qc/Win

where Qc is the heat removed from the cold reservoir (ie the inside of the fridge).

AM
 
  • #5
I meant Qout as in what you get out of doing the cycle. We always care about the ratio of what you get out compared with what you put in (e = Out/In). What you get out here is Qc, and what you put in is work W.
 

Similar threads

Back
Top