# Energy required to heat a house

by Runaway
Tags: carnot engine, cop, thermodynamics
 P: 48 1. The problem statement, all variables and given/known data A) A house loses thermal energy through the exterior walls and roof at a rate of 5440 W when the interior temperature is 23.4 C and the outside temperature is −6.6 C. Calculate the electric power required to maintain the interior temperature at Ti if the electric power is used in electric resistance heaters (which convert all of the electricity supplied to thermal energy). Answer in units of W. B) Find the electric power required to maintain the interior temperature at Ti if the electric power is used to operate the compressor of a heat pump with a coeﬃcient of performance equal to 0.3 times the Carnot cycle value. Answer in units of W. 2. Relevant equations COP = Qh / W 3. The attempt at a solution Part A is 5440 W because for the temperature to stay constant the energy leaving the house must be replaced at the same rate. I'm not sure how to set up an equation for part B.
HW Helper
P: 6,684
 Quote by Runaway COP = Qh / W .... I'm not sure how to set up an equation for part B.
What is the COP for a Carnot heat pump operating between these temperatures? What is .3 times that (actual COP)? Use that actual COP of the heat engine and the above equation to determine the amount of work (per second) required to deliver that same 5440 J of heat (per second).

AM
 P: 48 Cop = 296.55 k / (296.55 k - 266.55 k) cop = 9.89 (5440 w * 9.89*.3) = 16140.48w
HW Helper
P: 6,684
Energy required to heat a house

 Quote by Runaway Cop = 296.55 k / (296.55 k - 266.55 k) cop = 9.89 (5440 w * 9.89*.3) = 16140.48w
I thought you said COP = Qh/W. What is this 16140 W supposed to be?

AM
 P: 48 I used COP = Qh/(Qh-Qc) = Th/(Th-Tc) to find the COP then I took that and multiplied it by the output * .3 and got 16140 J/S, or in other words, 16140 Watts
HW Helper
P: 6,684
 Quote by Runaway I used COP = Qh/(Qh-Qc) = Th/(Th-Tc) to find the COP then I took that and multiplied it by the output * .3 and got 16140 J/S, or in other words, 16140 Watts
Yes. I see that. My question is why did you multiply COP by Qh? I thought you were supposed to find W.

AM
 P: 48 So if I follow you, the answer would be COP = Qh/W COP * W = Qh W=Qh/COP W= 5440 J/s/(9.89*.3) = 1833.5 J/s