Question

An ideally efficient air conditioner keeps the room air at 290 K when the outdoor air is at 305 K. How much work does it consume when delivering 900 J of heat outside?

Answer 1

44.26 J

Step-by-step explanation:

Efficiency=
`1-\frac {T_i}{T_h}` where in this case
`T_i` is 290 K and
`T_h` is 305 hence

Efficiency=
`1-\frac {290}{305}=0.04918`

We also know that efficiency is given by
`\frac {W}{Q_h}\\0.04918=\frac {W}{900}`

W=0.04918*900=44.26 J