# Energy problem dealing with heat extracted from cold air

1. Nov 1, 2013

### astru025

1. The problem statement, all variables and given/known data

A heat pump requires 385 W of electrical power to deliver heat to your house at a rate of 2410 J per second. How many joules of energy are extracted from the cold air outside each second?

2. Relevant equations

COP= Qc / W . This was the only equation I could find.

3. The attempt at a solution
Any help would be nice! I tried the above equation but I turned out to be incorrect.

2. Nov 1, 2013

### Kishlay

use this and find out the ans...

3. Nov 1, 2013

### haruspex

What do you think happens to the electrical energy used by the pump?

4. Nov 1, 2013

### Staff: Mentor

This is the same as an air conditioner that is cooling the outside and rejecting the heat to the inside. If the heat pump operates reversibly, then Q1 is removed from the outside at temperature T1, and Q2 is added to the inside at temperature T2. For a reversible process, such that ΔS of the surroundings = 0, $\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$. Since T1<T2, Q2>Q1. The difference between Q2 and Q1 is the work done by the heat pump.