Question

A Refrigerator absorbs 650KJ heat from low temperature heat reservoir and consumes work of 250 KJ. The C. O. P. of Refrigerator is​

Answer 1

Answer:

Partial conversion of heat received to mechanical work W. ... Refrigerators and heat pumps are reversed heat engines. ... A reversed heat engine absorbs 250 kJ of heat from a low temperature region and ...

Answer 2

Given:

Heat absorbed by refrigerator from low-temperature reservoir =650KJ

Work consumed by refrigerator = 250KJ.

Find:

C.O.P of the refrigerator

Solution:

C.O.P is known as the coefficient of performance for a refrigerator.

Heat absorbed by the refrigerator from the low-temperature reservoir is denoted by 'Q low'

The work done in each cycle is given by 'W'

Since the work is consumed therefore it will be (-W) = -250KJ

Then, C.O.P is given by the formula:  COP = Q low/(-W)

                                                              C.O.P = 650KJ / (-(-250))

                                                               C.O.P = 2.6

The C.O.P of the refrigerator is 2.6.