Question: A mixture of 250 gm of water and 200 gm of ice at 0<sup>0</sup>C is kept is calorimeter of water equ...

Question

A mixture of 250 gm of water and 200 gm of ice at 0⁰C is kept is calorimeter of water equivalent 50 gm. If 200 gm of steam at 100⁰C is passed through the mixture then the final amount of water in the mixture will be (Latent Heat of ice = 80 cal/gm, latent Heat of vaporisation of water = 540 cal/gm and specific heat of water = 1 cal/gm⁰C) –

Answer 1

Solution

Heat required by (ice +water + calorimeter) to raise upto 100⁰C

= [200 × 80] + [500 × 1 × 100]

̃ 16000 + 50000

̃ 66000 calorie

Let "m" gm of steam condenses,

66000 = m × 540

̃ m » 122 gm

So final amount of water = 450 + 122 = 572gm