If 10 ~g ofice is 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 $$\begin{aligned} H_1=m s \Delta T & =\frac{40}{1000} \times\left(4.2 \times 10^3\right) \times(15-0) \\ & =2520 \mathrm{~J} \end{aligned}$$ Heat required to convert $10 \mathrm{~g}$ ice into $10 \mathrm{~g}$ water at $0^{\circ} \mathrm{C}$ is $H_2=m L=\frac{10}{1000} \times\left(3.36 \times 10^5\right)=3360 \mathrm{~J}$ 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 $0^{\circ} \mathrm{C}$.