How much steam at `100^(@) C` is to be passed into water of mass 100g at `20^(@)c` to raise its temperature by `5^(@)C` ? (Latent heat of steam is 540 cal /g and specific heat of water is `1 cal // g^(@)C`)

Explanation

Heat lost by steam = heat gained by water
`m_(2) L_(2) + m_(2) S (100 - t) = m_(w) S ( t- 20)`
where `m_(s)` is the mass of steam, `L_(s)` is the latent heat of s s steam, sis the specific heat of water and `m_(w)` is the mass of water .
Here, `L_(s)` = 540 cal/g , s = 1 cal/`g^(0)` C , `m_(w)` = 100 g ,
t = 20 + 5 = `25^(@)` C
`m_(2) xx 540 + m_(2) xx 1 (100 - 25) = 100 xx 1 (25 - 20 )`
`m_(2) xx 615 = 500 therefore m_(2) = (500)/(615) = 0.813` g.