A heat pump is used to keep a house warm at 20°C. Assume ideal (Carnot) behavior. How much work is required of the pump to deliver 2970 J of heat into the house for each of the following outdoor temperatures?
(a) 0°C
(b) -15°C
Physics help please and thank you cant figure out where to start
Update: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
Answers & Comments
Verified answer
With heat pumps and refrigerators you never use heat balance equations.
Ideal Coefficient Of Performance (COP) for heating = Th / (Th - Tc), where Th is the high temperature, and Tc is the low temperature.
You can work ot the COP(heat) for the temperatures you have. Do not do that before convertion to Kelvin absolute units.
COP(heat) = heat delivered / work required -----> work required = heat delivered / COP(heat).
You can perform the calculation easily.
Good luck.
Actually, you don't need the insulation value for the house because the problem statement tells you how much heat you need to deliver (2970 J).
From your isentropic relation, Qc/Tc = Qh/Th, you can compute Qc given that Qh = 2970 J. You are given the two temperatures, just be sure to convert them to kelvin as you already said. The Qc that you calculate will be less than Qh. So the work needed is just the difference between the two as indicated by your other equation, Qc + W = Qh.
You need a value for insulation of the house or thermal heat transfer.