Answer to Question #83975 in Molecular Physics | Thermodynamics for Philip

Question

A 15.0-kg block of ice at 0.0°C melts to liquid water at 0.0°C inside a large room at 20.0°C. Treat the ice and the
room as an isolated system, and assume that the room is large enough for its temperature change to be ignored.
(Given that Lf = 3.34×10^5 J/kg for water)
a) Is the melting of the ice reversible or irreversible? Explain, using simple physical reasoning without resorting to
any equations.
b) Calculate the net entropy change of the system during this process. Explain whether or not this result is
consistent with your answer to part (a).

Answer 1

The melting of the ice inside a large room is irreversible process. This is because the heat flow from room (hot) to the ice (cold) is irreversible.

The net entropy change

∆S=∆S_ice+∆S_room

∆S_ice=∫〖δQ/T=Q/T_1 〗=(L_f m)/T_1 =(3.34×〖10〗^5 J/kg×15.0 kg)/(273 K)=18352 J/K

∆S_room=∫〖δQ/T=-Q/T_2 〗=-(3.34×〖10〗^5 J/kg×15.0 kg)/(293 K)=-17099 J/K

So

∆S=183526 J/K+(-17099 J/K)=1253 J/K >0

We obtain that net entropy increases. Therefore the melting of the ice inside a large room is irreversible process.