Question

(a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to another environment at 45.0ºC ? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per 3.60×106J (a kilowatt-hour)?

Answer 1

Answer:

a) 3.242

b) 1291.178 KJ

c) 3.59 cents

Explanation:

a) First all the temperature units should be in Kelvin. -30°C + 273.15 = 243.15K and 45°C + 273.15 = 318.15K. Coefficient of performance for refrigerator is COP = TL/(TH-TL) where TL is Lower Temperature and TH is the Higher Temperature. COP = 243.15/(318.15-243.15) = 3.242

b) COP  = Cooling Effect/Work Input. Cooling Effect (or Desired Output) is 4186 kJ.

So Work Input = Cooling Effect/COP = 4186KJ/3.242= 1291.178 KJ

c) 3.6 MJ (3600kJ) is for 10 cents, using ratio for cost: energy used

$\frac{10}{3600 kJ} =\frac{x}{1291.178kJ}$

3.59 cents