To what height can a 1 kg load be lifted due to the energy that is released during cooling to 0 degrees

To what height can a 1 kg load be lifted due to the energy that is released during cooling to 0 degrees Celsius of steam weighing 200 grams taken at 100 degrees Celsius.

mg = 1 kg.

g = 10 m / s2.

mp = 200 g = 0.2 kg.

t1 = 100 ° C.

t2 = 0 ° C.

С = 4200 J / kg * ° C.

λ = 2.3 * 106 J / kg * ° C.

Q = En.

h -?

According to the law of conservation of energy, the entire amount of heat Q, which is released during cooling of water vapor, turns into the potential energy of the load Ep: Q = Ep.

Q = λ * mp + C * mp * (t1 – t2).

Ep = mg * g * h.

λ * mp + C * mp * (t1 – t2) = mg * g * h.

h = (λ * mp + C * mp * (t1 – t2)) / mg * g.

h = (2.3 * 106 J / kg * ° C * 0.2 kg + 4200 J / kg * ° C * 0.2 kg * (100 ° C – 0 ° C) / 1 kg * 10 m / s2 = 54400 m.

Answer: the load will be able to rise to a height of h = 54400 m.