An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

Hot temperature TH= 300K

Cold temperature TL= 260K

Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

Wnet, in = QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

Thus, the 133.33J represents the initial work input during the heat transfer process.

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J