Question

5 g of ice at $ -30^\circ \text{C} $ and 20 g of water at $ 35^\circ \text{C} $ are mixed in a calorimeter. The final temperature of the mixture is:
(Neglect heat capacity of the calorimeter. Specific heat capacities: ice = $ 0.5 \, \text{cal/g}^\circ\text{C} $, water = $ 1 \, \text{cal/g}^\circ\text{C} $, latent heat of fusion of ice = $ 80 \, \text{cal/g} $)

• A. \( 0^\circ \text{C} \)
• B. \( 4^\circ \text{C} \)
• C. \( 5^\circ \text{C} \)
• D. \( 9^\circ \text{C} \)

Hint:

Use conservation of heat: Heat lost = Heat gained. Don't forget to include latent and specific heats.

Answer 1

The Correct Option is D

Heat required by ice to reach \( 0^\circ \text{C} \): \[ Q_1 = m c \Delta T = 5 \cdot 0.5 \cdot 30 = 75\, \text{cal} \] Heat required to melt ice: \[ Q_2 = 5 \cdot 80 = 400\, \text{cal} \] Total heat absorbed by ice: \[ Q_{total} = Q_1 + Q_2 = 75 + 400 = 475\, \text{cal} \] Heat released by water to reach \( T_f \): \[ Q = m c \Delta T = 20 \cdot 1 \cdot (35 - T_f) \] Equating: \[ 20(35 - T_f) = 475 \Rightarrow 700 - 20T_f = 475 \Rightarrow T_f = \frac{225}{20} = 11.25^\circ \text{C} \] But this means not all water is used in heating. The ice melts fully and extra heat raises the temperature of water formed from ice. So, total water mass = 25 g \[ Q_{rem} = 700 - 475 = 225\, \text{cal} \] Temperature rise: \[ \Delta T = \frac{Q}{mc} = \frac{225}{25 \cdot 1} = 9^\circ \text{C} \Rightarrow T_f = 0 + 9 = 9^\circ \text{C} \]