Question

A refrigerator what we call a heat engine running backwards for the purpose of making a cool place cooler. Refrigerators/freezers/air conditioners extract heat from the cold reservoir (the refrigerator or home interior) and deposit it in the hot reservoir (the kitchen or outdoors). However, we take advantage of the same physics/engineering principles to build heat pumps which are heat engines running backwards for the purpose of making a warm place warmer. They extract heat from the cold reservoir (the outdoors) and dump it in the hot reservoir (the indoors). You want to offset a 5000 W rate of heat loss through your floor and walls via conduction (averaged over day and night). (A) If you use an electric space heater, it will draw 5000 W of power, converting electric energy directly into heat. SRP’s basic winter rate is about 8 cents per kWhr. Find the cost to continuously run a space heater that will keep the house warm for one month. (B) It’s much smarter to use a heat pump – if it has a coefficient of performance of 5, find the electric power required and cost to run the heat pump continuously for one month.

I need help with this question, Especially with how to start it off. Thank You!

Answer 1

A) eletrical Energy consumed in one month, Q = P*t

= 5000*30*24*60*60

= 1.296*10^10 J

= 1.296*10^10/(3.6*10^6) kWh

= 3600 Kwh

cost = 3600*8

= 28800 cents

= 288 $ <<<<<<------------Answer

b)
COP of heat pump, COP = heat delivered to Warm place/Work done by the engine

5 = 5000/Work done by the engine

Workdone by the engine = 5000/5

= 1000 J/s

Energy used in one month, Q = 1000*30*24*60*60

= 2.592*10^9 J

= 2.592*10^9/(3.6*10^6) kWh

= 720 Kwh

cost = 720*8

= 5760 cents

= 57.6 $ <<<<<<------------Answer