Question

QH- QC

The heat pump illustrated beside uses 1.5 kW mechanical power (that's is 1500 J/s) and extracts 1.2 million calories per hour from the outside air.

• A) How much heat is pumped into the house every day?

• B) If this heat pump is made such that it is ideal and reversible, and you decide to heat your house with a gas furnace and use the heat pump as an heat engine to produce mechanical work, then what would be its efficiency?

• C) If the outside temperature is -15 deg. C, what temperature would you have to set indoors?

Answer 1

Qh = Qc + W

Heat extracted from outside per hour = 1.2×10^6 calories

Heat extracted in 24 hrs, Qc = 2.9×10^7 calories

Work done per second = 1500 Joules

Work done per day, W = 1500*86400(seconds per day)/4.18(joules per calorie) = 3.1×10^7 calories

Hence heat pumped into house per day = (2.9+3.1)×10^7

= 6.0×10^7 calories

b) Efficiency of engine = W*100/Qh

= 3.1×10^7* (100) / 6×10^7 = 51.7%

c) For idea engine or heat pump,

Qh / Qc = Th/Tc. , where temperatures are in Kelvin. Hence

6/2.9 = Th/258. (-15 deg C = -15+273 = 258 K)

Th = 534 K = 261 deg C