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Q.
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 = $4.2 \times 10^3\, J\, kg^{-1} K^{-1}$, Latent heat of fusion of ice = $3.36 \times 10^5 \,J \, kg^{-1})$
Heat lost by water to come from 15 $^{\circ}$C to 0 $^{\circ}$C is
$H_{1}=ms\Delta T=\frac{40}{1000}\times\left(4.2\times10^{3}\right)\times\left(15-0\right)$
$\quad$ $\quad\quad\quad\quad$ = 2520 J
Heat required to convert 10 g ice into 10 g water at 0$^{\circ}$C is
$H_{2}=mL=\frac{10}{1000}\times\left(3.36\times10^{5}\right)=3360$ J
Since H2 > H1, so the whole ice will not be converted into water, whereas the temperature of the whole water will be 0$^{\circ}$C.
Therefore the temperature of the mixture is 0$^{\circ}$C.