A mixture of 250 g of water and 200 g of ice at 0° C is kept in a calorimeter of water equivalent 50g . If 200g 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 cal 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 50g . If 200g 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 cal g- 1/° C ) (A) 450 g (B) 622 g (C) 572 g (D) 650 g

Tardigrade Admin · Accepted Answer

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

$450 g$

$622 g$

$572 g$

$650 g$

450g

622g

572g

650g

$450 g$

$622 g$

$572 g$

$650 g$