Question

a.) What is the best possible coefficient of performance for a refrigerator that cools an environment at -33.0 °C and exhausts heat to another environment at 39.0 °C? b.) How much work in joules must be done for a heat transfer of 3.925 × 106 J from the cold environment? c.)What is the cost of doing this, if the work costs 17.5 cents per 3.60 × 106 J (a kilowatt-hour)? d.) How many joules of heat transfer occurs into the warm environment?

Answer 1

SOLUTION :

We have,

Tc = -33 oC = 240 K

Th = 39 oC = 312 K

(a)

Formula for coefficient of performance for a refrigerator,

COP_ref = Tc / Th - Tc

COP_ref = 240 / (312 - 240)

COP_ref = 3.33

(b)

Work done  for a heat transfer ,

W = Q / COP_ref

W = 3.925 × 10^6 / 3.33

W = 1.17*10^6 J

(c)

cost = (17.5 / 3.6*10^6) * 1.17*10^6

cost = 5.7 cents

(d)

heat transfer into the warm environment,

Qh = Q + W

Qh =  3.925 × 10^6 + 1.17*10^6

Qh = 5.1*10^6 J