Question

The whole mixture in the calorimeter becomes ice if

• A. C₁t₂ + C₂t₂ + L + C₃t₃ > 0
• B. C₁t₂ + C₂t₂ + L + C₃t₃ < 0
• C. C₁t₂ + C₂t₂ - L + C₃t₃ > 0
• D. C₁t₂ + C₂t₂ - L - C₃t₃ < 0

Answer 1

Answer: (C)

C₁t₂ + C₂t₂ - L + C₃t₃ > 0

Answer 2

For the whole mixture to become ice, the heat released by the calorimeter and water as they cool to $0^\circ C$ must be sufficient to warm the ice to $0^\circ C$ and then freeze all the water. Heat released by calorimeter cooling from $t_2$ to $0^\circ C$: $Q_{cal} = m C_1 (t_2 - 0) = m C_1 t_2$. Heat released by water cooling from $t_2$ to $0^\circ C$: $Q_{water} = m C_2 (t_2 - 0) = m C_2 t_2$. Heat gained by ice to warm from $t_1$ to $0^\circ C$: $Q_{ice\_warm} = m C_3 (0 - t_1) = -m C_3 t_1$. Heat required to freeze the water at $0^\circ C$: $Q_{freeze\_water} = m L$.

The condition for the entire mixture to become ice is that the total heat released is greater than or equal to the total heat required: $Q_{cal} + Q_{water} \ge Q_{ice\_warm} + Q_{freeze\_water}$ $m C_1 t_2 + m C_2 t_2 \ge -m C_3 t_1 + m L$

Assuming $m > 0$, we divide by $m$: $C_1 t_2 + C_2 t_2 \ge -C_3 t_1 + L$ $C_1 t_2 + C_2 t_2 + C_3 t_1 - L \ge 0$

The options provided use $t_3$ instead of $t_1$. Assuming $t_3$ is a typo for $t_1$, the condition becomes $C_1 t_2 + C_2 t_2 + C_3 t_1 - L \ge 0$. Option (C) is $C_1 t_2 + C_2 t_2 - L + C_3 t_1 > 0$, which matches the derived condition for the mixture to become ice with a final temperature below $0^\circ C$.