If 10 ~g oficeis added to 40 ~g of water at 15^ C , then the...

Heat lost by water to come from $15^{{}^{\circ}} \mathrm{C}$ to $0^{{}^{\circ}} \mathrm{C}$ is $$\begin{aligned} H_1=m s \Delta T & =\frac{40}{1000} \times \left(4.2 \times10^3\right)\times(15-0) \\ & =2520 \mathrm{ J} \end{aligned}$$Heat required$to$convert 10 \mathrm{ g} ice into 10 \mathrm{ g} water at 0^{{}^{\circ}} \mathrm{C} is H_2=mL=\frac{10}{1000} \times\left(3.36 \times10^5\right)=3360 \mathrm{ J}Since$H_2 > H_1, so the whole ice will not be converted into water, whereas the$temperature of the whole water will be 0^{{}^{\circ}}\mathrm{C}. Therefore$the$temperature of the mixture is 0^{{}^{\circ}}\mathrm{C}.$