Question

Two experiments P and Q are performed with water in various states of phases. Both experiments are performed at constant pressures. Heat is supplied to the systems at the same constant rate in both experiments. The latent heat of fusion, vaporization, and sublimation of water are respectively $334kJ/kg$, $2501kJ/kg$, and $2834kJ/kg$. The phase diagram of water and the temperature-time schematic plots of the two experiments are depicted below (bold curve for P and dashed curve for Q). Then one can conclude that:

• A. the initial mass of water in P is equal to that in Q.
• B. at time $t_o$. P is in a pure phase while Q is in a mixed phase.
• C. experiment P has been performed at a pressure higher than Q.
• D. at $t>t_o$, Q goes to the liquid state.

Answer 1

Answer: (B), (C)

The correct options are
B at time $t_o$. P is in a pure phase while Q is in a mixed phase.
C experiment P has been performed at a pressure higher than Q.

In the graph initial phase is ice as temperature is negative initially and the horizontal line near t-axis is the process of fusion of ice.

Since, we see that fusion for $Q$ takes more time and hence more heat than$P$, hence, mass of water in $Q$ should be greater than $P$.

At $t=t_o$, $P$ has completely converted into liquid and is in the pure state. And $Q$ is in transition phase and hence in mixed phase.

Since, fusion part of P is slightly above Q , hence pressure is higher for P.

Hence, (B) and (C) are correct options.