Question

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. (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).

Short answer

The melting of ice is irreversible due to increase in entropy of the system.

Step-by-step solution

Identification of given data

The temperature of ice is\({T_1} = 0\;^\circ {\rm{C}}\)

The temperature of room is\({T_2} = 20\;^\circ {\rm{C}}\)

The mass of ice block is \(m = 15\;{\rm{kg}}\)

Conceptual Explanation

The process is reversible and possible if the entropy of the system decreases otherwise process will be irreversible.

Whether the process is reversible or irreversible

The entropy of a system is the molecular disorder of molecules of the system. This disorder of the system is increased when the molecules are disturbed from its equilibrium state.

When the ice is converted into water then molecular disorder increases because molecules can move more randomly in liquid state. When temperature of water increases and reaches to room temperature then molecular disorder of water molecules also increases.

The entropy of the system increases due to net increase in molecular disorder so the melting of ice is irreversible.

Therefore, the melting of ice is irreversible due to increase in entropy of the system.