A 3.00 watt electric motor is plugged into an electrical outlet. It takes the motor 30.00 seconds to lift a mass of 254.9 g a distance of 10.00 cm. In that time, the motor has used 90.00 J of energy. Assuming no energy leaves the system, how much heat has been added to the system by the end of those 30 seconds?

Answer 1

the lifted mass gains potential energy ... m * g * h
... .2549 kg * 9.8 m/s^2 * .01000 m = ?

the rest of the 90.00 J consumed by the motor becomes "waste" heat

Answer 2

power = work/time, so
work = power*time = 3.00*30.00 = 90.00 J

but the actual work done was only
work = Fd = 0.2549*9.8*0.10 = 0.2498J

so, what do you think?