Calculating Heat Expelled from a Heat Pump

[jdog6]

A heat pump has a coefficient of performance of 7.05. If the heat pump absorbs 20 cal of heat from the cold outdoors in each cycle, find the heat expelled to the warm indoors. Answer in units of cal.
I believe COP (heating mode) : 7.05 and COP = Qh/W
Qc = 20 cal
so I have to find Qh= ? cal
I don't know an equation to put all this together?
Please help, thank you.

 

[lightgrav]

Is Energy conserved?

If the coeff. of performance is 7.05,
how much Work is done in one cycle?

 

[jdog6]

the question does not say all i have is what i posted

 

[lightgrav]

The entire subject of Thermodynamics
is FOUNDED on Conservation of Energy
(Energy can be moved around, but not created/destroyed).

So, YES, Energy is conserved.

Where do you think the Energy comes from,
that is dumped into the (warm) room?

 

[jdog6]

The energy comes from rods outside.

 

[lightgrav]

Not all of it ... it takes *Work* to operate the pump!
Qh/W = 7.05 , so you have to PAY for the Energy
(electricity, probably) for the W = Qh/7 of the heat.

So, what fraction do you get "for free", from outside?

 

[philrainey]

Qh=W+Qc so W=(W+20)/7.05
7.05W-W=20
W(7.05-1)=20
6.05W=20
W=20/6.05
W=3.306
Qh=3.306+20
Qh=23.306 cal/cycle
and to check it 23.306/3.306=7.05