Question

10 g of ice is added in 40 g of water at `15^@C.` Calculate the temperature of the mixture. Sp. Heat of water ` = 4.2 xx 10^3 J kg^(-1) K^(-1), ` Latent of fusion of ice ` = 3.36 xx 10^5 J kg^(-1)`

Answer 1

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.`