(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)?

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

[tex]\frac{10}{3600 kJ} =\frac{x}{1291.178kJ}[/tex]

3.59 cents

andromache andromache · Answer 2 · 2022-02-08T14:59:52+01:00

a) The best coefficient of performance for a refrigerator is 3.242

b)The number of work in joules is 1291.178 KJ

c) The cost of doing this is 3.59 cents

Calculation of the best performance coefficient, number of work, and the cost:

a) The best coefficient should be

But before that the temperature is

= -30°C + 273.15 = 243.15K and 45°C + 273.15 = 318.15K.

Now

Coefficient of performance for refrigerator is COP

[tex]= TL\div (TH-TL) \\\\ = 243.15\div (318.15-243.15)[/tex]

= 3.242

b)

Since

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

So Work Input =[tex]Cooling Effect\div COP = 4186KJ\div 3.242[/tex]= 1291.178 KJ

c) Now the cost is

[tex]10 \div 3600J = x\div 1291.178KJ[/tex]

x = 3.59 cents

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