If 10 g of ice is added to 40 g of water at `15^(@)C`, then the temperature of the mixture is (specific heat of water `= 4.2 xx 10^(3) j kg^(-1) K^(-1)`, Latent heat of fusion of ice `= 3.36 xx 10^(5) j kg^(-1)`)

Explanation

Heat lost by water to come from `15^@C` to `0^@C is H_1 = (40)/(1000) xx(4.2xx 10^3) xx(15 - 0) = 2520 J` Heat required to convert 10 g ice into 10 g water at `0^@C is H_2 = (10)/(1000) xx (3.36 xx 10^5) = 3360 J` Since `H_2 gt H_1` so the whole ice will not be converted into water, whereas the temperature of the whole water will be `0^@C.` Therefore the temperature of the mixture is `0^@C.`