A well insulated calorimeter contains a mixture of water and ice. A heater supplies energy at the rate of 30 watt to the mixture. At the ends of second, third and fourth minute after heater is switched on, temperature of the mixture is $0 ° C$ , $2 ° C$ and $8 ° C$ respectively. How many grams of ice were initially present in the calorimeter? (Latent heat of melting of ice = 400 J/g and Specific heat of water $= 4 J / g ° C$ )

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answer is 12.

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Detailed Solution

Let x be the amount of water and y be the amount of ice at the end of second minute
${yL}_{ice} + (x + y) C_{water} ΔT = 30 \times 60$
$400 y + (x + y) 4 \times 2 = 1800$ ……………….(1)
$(x + y) \times 4 \times 6 = 30 \times 60$ …………(2)
Solving (i) and (ii)
$400 y + \frac{1800}{6} \times 2 = 1800, 400 y = 1200 \Rightarrow y = 3 g$
Initial amount of ice $= \frac{30 \times 120}{400} + 3 = 12 g$