Question

Ice at $-20^{\circ} C$ os added tp $50\, g$ of water at $40^{\circ}C$. When the temperature of the mixture reaches $0^{\circ}C$, it is found that $20\, g$ of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = $4.2 \; J/g/^{\circ}C$) Specific heat of Ice = $2.1 \; J/g/^{\circ}C$ Heat of fusion of water at $0^{\circ}C = 334 \; J/g$)

• A. 50 g
• B. 40 g
• C. 60 g
• D. 100 g

Answer 1

Answer: (B)

40 g

Answer 2

Let amount of ice is m gm.
According to principal of calorimeter heat taken by ice = heat given by water
$\therefore \; 20 \times 2.1 \times m + (m -20 ) \times 334$
$= 50 \times 4.2 \times 40$
376 m = 8400 + 6680
$m = 40.1$