Question

#10: A heat pump requires 430 W of electrical power to deliver heat to your house at a rate of 2390 J per second. How many joules of energy are extracted from the cold air outside each second?

#11: The coefficient of performance for a heat pump is defined as the ratio of the heat delivered to the warmer system to the work required. If a heat pump requires an input of 360 W of electrical energy and has a coefficient of performance of 3.0, how much energy is delivered to the inside of the house each second?

Answer 1

10. Heat extracted by pump per second = heat to be supplied to the house - power of the pump

= 2390 J/s - 430 J/s = 1960 J/s

11. Coefficient of performance = Energy delivered per second by the pump (J/s) / power of the pump

or, 3 = Energy delivered per second by the pump (J/s) /360 W

or, Energy delivered per second by the pump (J/s) = 360 * 3 = 1080 J/s