If 10 ~g ofice is added to 40 ~g of water at 15^ C , then the temperature of the mixture is (specific heat of water =4.2 10^3 ~J ~kg ⁻¹ ~K ⁻¹ , Latent heat of fusion of ice =3.36 10^5 ~J ~kg ⁻¹ )

Step-by-step solution

Heat lost by water to come from 15^ C to 0^ C is aligned H₁=m s T & = 40 1000 (4.2 10^3 ) (15-0) & =2520 ~J aligned Heat required to convert 10 ~g ice into 10 ~g water at 0^ C is H₂=mL= 10 1000 (3.36 10^5 )=3360 ~J Since H₂ > H₁ , 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 .