(a) What is the best coefficient of performance for a heat pump that has a hot reservoir temperature of 50.0ºC and a cold reservoir temperature of −20.0ºC ? (b) How much heat transfer occurs into the warm environment if 3.60×107 J of work ( 10.0kW ⋅ h ) is put into it? (c) If the cost of this work input is 10.0 cents/kW ⋅ h , how does its cost compare with the direct heat transfer achieved by burning natural gas at a cost of 85.0 cents per therm. (A therm is a common unit of energy for natural gas and equals 1.055×108 J .)

Solution 41PE

a)

COPhp = Th / Th - Tc

COPhp = (50 OC +273.15 ) K / (50 OC +273.15 ) K - (50 OC +273.15 ) K

= 232.15 K / 70.0 K

= 4.16 K

the coefficient of performance of the heat engine is COPhp = 4.16 K

b)

Qh = (4.614)(3.60x107J)

= (1.661x108J)(1 kal / 4186 J)

= 3.97x104 kcal

he heat transfer into the warm environment is 3.97x104 kcal

c)

the heat pump is more than three times (1 / 0.29) as expensive as natural gas.