Ice at -20° C os added tp 50 g of water at 40°C. When the temperature of the mixture reaches 0°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/°C) Specific heat of Ice = 2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g)

Question

Ice at -20° C os added tp 50 g of water at 40°C. When the temperature of the mixture reaches 0°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/°C) Specific heat of Ice = 2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g) (A) 50 g (B) 40 g (C) 60 g (D) 100 g

Tardigrade Admin · Accepted Answer

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

50 g

40 g

60 g

100 g

Let amount of ice is m gm. According to principal of calorimeter heat taken by ice = heat given by water ∴20×2.1×m+(m−20)×334 =50×4.2×40 376 m = 8400 + 6680 m=40.1

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