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Question
Calculate the mass of ice required to lower the temperature of 300 g of water 40°C to water at 0°C.
(Specific latent heat of ice = 336 J/g, the Specific heat capacity of water = 4.2J/g°C)
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Solution
Let m be the mass of the ice to be added.
Heat energy required to melt to lower the temperature is = mL = m x 336
Heat energy imparted by the water in fall of its temperature from 40°C to
0°C = mass of the water x specific heat capacity x fall in temperature
= 300 x 4.2 x 40°C
If there is no loss of heat,
m x 336 J/g = 300 g x 4.2 J/g°C x 40°C
`:. m = (300 xx 4.2 xx 40)/336`
∴ m = 150g
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