Question

10 g of ice at $-20^\circ$C is dropped into a calorimeter of water equivalent 20 g, containing 10 g of water at $20^\circ$C. At equilibrium, the mass of the ice in the mixture is

• A. 3.75 g
• B. 20 g
• C. 3.33 g
• D. 5 g

Hint:

Always account for heating of ice to 0°C, latent heat of fusion, and calorimeter effect.

Answer 1

The Correct Option is A

1. Heat lost by water and calorimeter = Heat gained by ice to reach 0°C + heat to melt + remaining ice mass
2. Let $m$ = mass of ice melted.
3. $m \cdot 80 + m \cdot 0.42 = 10 \cdot 4.2 \cdot 20 + 20 \cdot 4.2 \cdot 20$ (in Joules)
4. Solve for $m$ → $m = 3.75$ g