Q.

Ice at 20°C is added to 50g of water at 40°C . When the temperature of the mixture reaches 0°C , it is found that 20g of ice still unmelted. The amount of ice added to the water was close to 
(Take, specific heat of water = 4.2J/g/°C , specific heat of ice =2.1J/g/°C and latent heat of fusion of water at 0°C=334J/g  )

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a

100g

b

60g

c

40g

d

50g

answer is A.

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Detailed Solution

Let amount of the ice be  x gm, according to the principle of calorimeter,
Heat lost by water = heat gained by ice , Here, heat lost by water, ΔQ=mSwaterΔT 
Subtracting the given values, we get  ΔQ=50×4.2×40=8400
Heat given by ice, ΔQ=xSiceΔT+x20L 
42x+334x6680=8400x=15080376=40.1g40g

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