Ice at -20°C is added to 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 = 4.2J/g/°C Specific heat of Ice = 2.1 J/g/°C. Heat of fusion of water at 0°C = 334 J/g
Text Solution
Verified by ExpertsThe correct answer is:
B
Heat lost by water Δ Q = ms Δ T = 50× 1 × 40 = 2000 cal
Let mass of ice = m
Heat gain by ice Δ Q = m × 
× 20 + (m – 20) × 80
= 10 m + 80 m – 1600 = 90 m – 1600
Heat gain = Heat lost
90 m – 1600 = 2000
m = 40 gm
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