Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°c, and is compressed at steady state to 12 bar, 80°c. the volumetric flow rate of the refrigerant entering is 4 m3/min. the work input to the compressor is 60 kj per kg of refrigerant flowing. neglecting kinetic and potential energy effects, determine the magnitude of the heat transfer rate from the compressor, in kw.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j