300 grams of water at ${25}^{\circ }\text{C}$  is added to 100 grams of ice at $0^{\circ }\text{C}$ , the final temperature of the mixture is 

Water
$$\begin{gathered} {\text{amount of heat lost by water is dQ}}_{\text{1}}\text{=mSdθ} \\ here m is mass of water ; S is specific heat of water; \mathrm{d\theta } is change in temperature of water \\ when water looses its temperature from {25}^{0}C to 0^{0}C, amount of heat lost is {\mathrm{dQ}}_{\text{1}} \end{gathered}$$
${\mathrm{dQ}}_{\text{1}}\text{=300}\times \text{1}\times \text{25=7500\hspace{0.17em}\hspace{0.17em}calorie}$
For Ice
$$\begin{gathered} {\text{amount of heat required to convert from ice to water is dQ}}_2\text{=mL} \\ where m is mass of ice ; L = latent heat of fusion of ice=80 cal g^{-1} \end{gathered}$$
${\mathrm{dQ}}_2\text{=100}\times 80=8000\text{\hspace{0.17em}\hspace{0.17em}cal}$
${\text{dQ}}_2{\text{>dQ}}_1\text{i.e., Heat required is greater than heat available}$
Still ice remains= ${\text{0}}^{\text{o}}\text{C}$