ACME Refrigerator Co has claimed that a new model freezer op

ACME Refrigerator Co. has claimed that a new model freezer operates using 310 W of power, has an internal temperature of 260 K (TL), and rejects 510 W (QH) of heat to an environment at 635 K (TH).

1. Calculate the heat rate (QL) that the refrigerator receives from the cold storage.

2. Calculate the coefficient of performance (COP) of this refrigerator.

3. Calculate the ideal COP of a refrigerator operating between the given TH and TL.

4. Is the manufacturer’s claim valid? Why or why not?

Solution

>> Now, Tl = 260 K

Th = 635 K

Qh = 510 W = Heat supplied to higher temperature body

and, W = Power required = 310 W

So, Heat Rate, Ql = Qh - W = 510 - 310 = 200 W ......Heat refrigerator recieves fro the storage.....

2). As, COP = Ql/W = 200/310 = 0.645

3). Ideally, COP = Ql/W = Tl/(Th - Tl) = 260/(635 - 260) = 0.6933  

4). As, COP of given refrigerator is less than Maximum COP (or Canot Refrigerator COP), So, manufacturer\'s claim is valid .........