How much energy does it extract from the outside air with coefficient of performance 3.85?

A heat pump has a coefficient of performance of 3.85 and operates with a power consumption of 6.91 x 10^3 W.

The first part of the question asked How much energy does it deliver into a home during 8 h of continuous operation? and I got 766180800 J.

I need help on the last part which asks How much energy does it extract from the outside air? answer in Joules.

Qout = Qin + Win

Qin = Qout - Win
= Qout - Qout/3.85
= 0.7403 Qout

Use your previously caclulated value of Qout, which was correct. Scientific notation would be preferable

I got lost when you got to:

= Qout - Qout/3.85
= 0.7403 Qout

What did you exactly do? I'm not seeing where those two Qout's came from and how you got the .7403.

To determine the amount of energy extracted from the outside air, we can use the coefficient of performance (COP) of the heat pump.

The equation for calculating the output energy (Qh) is:
Qh = COP * input energy (Qc)

Given that the power consumption of the heat pump is 6.91 x 10^3 W and the COP is 3.85, we can calculate the input energy (Qc) using the formula:
Qc = Power * time

Qc = (6.91 x 10^3 W) * (8 h) = 55,280 J

To find the output energy (Qh), we multiply the input energy (Qc) with the COP:
Qh = 3.85 * 55,280 J = 212,708 J

Therefore, the heat pump extracts 212,708 Joules of energy from the outside air.

To determine how much energy the heat pump extracts from the outside air, you can use the coefficient of performance (COP) and the power consumption of the heat pump.

The COP is defined as the ratio of the heat output (energy delivered) to the work input (power consumed). In this case, the COP is given as 3.85.

The formula to calculate the heat delivered by the heat pump is:

Heat delivered = COP * Power consumed

Substituting the given values:

Heat delivered = 3.85 * 6.91 x 10^3 W

Now we can calculate the heat delivered in joules (J). To do this, we need to convert the power consumption from watts to joules per second (J/s) by multiplying it by the duration (8 hours). Then, since 1 watt-second (W·s) is equal to 1 joule (J), we can simply multiply the value by 3600 (to convert hours to seconds).

Heat delivered = (3.85 * 6.91 x 10^3 W) * (8 hours * 3600 seconds/hour)

By evaluating this expression, we can find the energy extracted from the outside air in joules.