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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 = $4.2 J / g / ° C$ , specific heat of ice = $2.1 J / g / ° C$ and latent heat of fusion of water at $0 ° C = 334 J / g$ )

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a

40g

b

50g

c

60g

d

100g

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, $Δ Q = m S_{w a t e r} Δ T$
Subtracting the given values, we get $Δ Q = 50 \times 4.2 \times 40 = 8400$
Heat given by ice, $Δ Q = x S_{i c e} Δ T + (x - 20) L$
$42 x + 334 x - 6680 = 8400 \Rightarrow x = \frac{15080}{376} = 40.1 g ≃ 40 g$

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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 )