A heat pump is used to keep a house warm at 22? C . Part A How much work is requ
ID: 1297068 • Letter: A
Question
A heat pump is used to keep a house warm at 22?C.
Part A
How much work is required of the pump to deliver 2800J of heat into the house if the outdoor temperature is 0?C. Assume a COP of 3.0
Part B
How much work is required of the pump to deliver 2800J of heat into the house if the outdoor temperature is -15?C? Assume a COP of 3.0
A heat pump is used to keep a house warm at 22?C.
Part A
How much work is required of the pump to deliver 2800J of heat into the house if the outdoor temperature is 0?C. Assume a COP of 3.0
Part B
How much work is required of the pump to deliver 2800J of heat into the house if the outdoor temperature is -15?C? Assume a COP of 3.0
Explanation / Answer
Conservation of energy (=1st law of thermodynamics) requires that energy transferred to the heat pump is equal the the amount of energy rejected by the pump.That means that heat absorbed by surrounding plus work done equals to the heat delivered to the house:
Qc + W = Qh
Carnot behavior means that heat pump operates without entropy production. So the entropy transferred with the absorbed heat is equal to entropy rejected with the heat into the house. (reversible work does not change entropy!)
?Sc = ?Sh
<=>
Qc/Tc = Qh/Th
where T stands for absolute temperatures in Kelvins
You have specified Qh and Th,Tc and want to know W. So substitute
Qc in energy expression using entropy relatiion. Then solve for W
Qc/Tc = Qh/Th
=>
Qc = Qh ? (Tc/Th)
Hence:
Qc + W = Qh
<=>
Qh?(Tc/Th) + W = Qh
=>
W = Qh ? [1 - (Tc/Th)]
a)
Qh = 2800J
Th = 22