Saving Electrical Energy with a Heat Pump: 75%

[M9501]

Homework Statement

The electrical energy supplied to a heater can be turned with 100% efficiency into thermal energy (room heating).Assume you use a heat pump with a COP of 4 instead to reach the same temperature
increase in the room.How much electrical energy will you save ?

Homework Equations

COP(heating)=Qh/W=Qc+W/W;

The Attempt at a Solution

Firstly i thought about getting the percentage of energy used which is equal to the work and then getting the percentage of energy saved by subtraction (100%-%ofUsedEnergy).

since temperature is constant then Qh is constant(am i correct?)
W=1/4 Qh (i guess something is wrong here)
∴the percentage of energy used=100*1/4=25%

∴the percentage of energy saved=75%

 

[rude man]

Homework Statement

The electrical energy supplied to a heater can be turned with 100% efficiency into thermal energy (room heating).Assume you use a heat pump with a COP of 4 instead to reach the same temperature
increase in the room.How much electrical energy will you save ?

Homework Equations

COP(heating)=Qh/W=(Qc+W)/W;

The Attempt at a Solution

Firstly i thought about getting the percentage of energy used which is equal to the work and then getting the percentage of energy saved by subtraction (100%-%ofUsedEnergy).

since temperature is constant then Qh is constant(am i correct?)
W=1/4 Qh (i guess something is wrong here)

Why?

the percentage of energy saved=75%

Your conclusion is correct. (That's why I run a heat pump myself! Unfortunately, if Tc drops too far then the Carnot efficiency drops: Qh/W = Th/(Th - Tc) so for the same Th the colder Tc the lower the COP. Of course, refrigerators in general do not use a Carnot cycle. Part of the cycle is via a throttling process {"expansion valve"} which is an irreversible part of the refrigeration cycle.)