Two 50 gm ice cubes are dropped into 250 gm of water into a glass. If the water was initially at a temperature of $25°C$  and the temperature of ice is $-15°C$. Find the final temperature (in K) of the water. 

(specific heat of $ice=0.5cal/gm/°Cand{L}_f=80cal/gm$).

Heat absorbed by ice to get converted from $-{15}^{\circ }\text{C}$ to $0^{0}C$ is 

${\text{Q}}_{\text{ice }}=100\text{*}0.5\text{*}15=750\text{cal}$

Thus temp of water will decrease to $25-3={22}^{\circ }\text{C}$ To
convert 100gms of ice at  $0^{0}C$ to water at $0^{0}C$ amount
of heat required is  $100\text{*}80=8000\text{cal}$ where  $\text{L}=80{\text{calgm}}^1$
is used.
To convert 250gm of water at ${22}^{0}C$  to water at$0^{0}C$ , the amount of heat released will be $250\text{*}1\text{*}22=5500\text{cal}$
, total heat released is less than the heat required to convert ice $0^{0}C$ to water at  $0^{0}C$
$\therefore$ aportion of the ice will get converted from ice at  ${\mathrm{o}}^{\wedge }\mathrm{\setminus }:{\text{Ctowaterato}}^{\wedge }\mathrm{O}\mathrm{\setminus }:\mathrm{C}$

. The amount ice converted from ice $0^{\wedge }\wedge :\mathrm{C}{\text{to water a to}}^{\wedge }\mathrm{O}\mathrm{\setminus }:\mathrm{C}\text{is}\frac{5500}{80}=\frac{550}{8}\mathrm{gms}$

Amount of ice remaining at $0^{0}C$ will be $(100-\frac{550}{8})=\frac{250}{8}=\frac{125}{4}\text{gms}$
Amount of water at $0^{0}C$  will be $(100-\frac{125}{4}+250)=\frac{1275}{4}\text{gms}$

Final temp of the mixture is $0^{0}C$