20g of steam at `100^(@)`C is passed into 100g of ice at `0^(0)`C. Find the resultant temperature if specific latent heat of steam is 540 callg., specific latent heat of ice is 80 cal/g and specific heat of water is 1 cal/`g^(0)` C .

Explanation

Let the resultant temperature of the mixture be `t^(0)` C. In the method of mixture,
heat lost by steam = heat gained by ice
`m_(2) L_(s) + m_(s) s (100 - t) = m_(c) L_(c) + m_(c) s (t - 0)`
where m is the mass of steam, `L_(s) `is the latent heat of steam, s is the specific heat of water, `m_(c)` is the mass of ice and `L_(c)` is the latent heat of ice.
Here, `m_(s) = 20 g , L_(s) = 540 `cal/g , s
1 cal/`g^(0) C , m_(c) = 100 g , L_(c) = 80 ` cal/g
`20 xx 540 + 20 xx 1 (100 - t) = 100 xx 80 + 100 xx 1 xx t`
`10800 + 2000 - 20t = 8000 + 100 t `
120 t = 4800
t = `40^(0)` C .