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

Latent heat of steam, `L_(s)=540 cal//g` Specific heat of water, `L_(w)=1 cal//g^(@)C` Mass of water , `m_(w)=100g` Accourding to method of mixture or from the principle of calorimetry we can write, Heat lost by steam = heat gained by water `i.e., m_(s) L_(s) +m_(s) S_(w)(100-t)= m_(w) S_(w) (t-20)`
`therefore m_(s)xx540 +mxx1(100-25)`
`rArr 100xx1xx(25-20)`
`rArr 615m_(s) =500(or) m_(s) = (500)/(615)= 0.813g`