silly question...!!?

Question

1/ A heat engine operating between temperatures of 500K and 200K is used to lift a 10-kg mass vertically at a constant speed of 5m/s.
(a) Determine the power the engine must supply ro lift the mass (hint: mechanical power).
(b) Determine the maximum possible efficiency at which the engine can operate.
(c) If the engine were to operate at the maximum possible efficiency, determine the following...
i) The rate at which the hot reservoir supplies heat to the engine
ii) The rate at which heat is exhausted to the cold resevoir.

Answer 1

Silly question? You must be kidding.

(a) Power is product of force applied to the object and it's velocity:
P=Fv=w v= mg v
P= 10 x 9.81x 5=490 Watts

(b) For an ideal heat engine (Carnot ) efficiency is

n= 1- T(cold) / T(hot)
n= 1 - 200/500= 0.6 0r 60 %

(c) Prequired was computed to 490 W and the engine is 60% efficient.
i) the rate at which the hot reservoir supplies heat to the engine
dQ(hot) /dt = P(out)/n= 490/.6= 817 W
similarly
dQ(cold) /dt = P(hot)(1- n) = 817 (1-.6)= 327 W

Perhaps it is a silly answer? :-)