A mixture of 250 g of water and 200 g of ice at 0 ° C is kept in a calorimeter of water equivalent 50 g . If 200 g of steam at 100 ° C is passed through the mixture then the final amount of water in the mixture will be (Latent Heat of ice = 80 cal g - 1 , Latent Heat of vaporisation of water = 540 cal g - 1 and specific heat of water = 1 c a l g - 1 / ° C )

Question

A mixture of 250 g of water and 200 g of ice at 0 ° C is kept in a calorimeter of water equivalent 50 g . If 200 g of steam at 100 ° C is passed through the mixture then the final amount of water in the mixture will be (Latent Heat of ice = 80 cal g - 1 , Latent Heat of vaporisation of water = 540 cal g - 1 and specific heat of water = 1 c a l g - 1 / ° C ), 450 g, 622 g, 572 g, 650 g

MARKS App · Accepted Answer

Heat required by ice to melt = 200 × 80 = 16000 cal Heat required by 250 g water (ice + initial water + calorimeter) to raise up to 100 ° C = 500 × 1 × 100 = 50000 cal ⇒ Total heat required = 16000 + 50000 = 66000 cal Let m g of steam condenses, then 66000 = m × 540 ⇒ m ≈ 122 g So final amount of water = 450 + 122 = 572 g

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