Question

Two $50gm$ ice cubes are dropped into $250gm$ of water into a glass. If the water was initially at a temperature of ${25}^{\circ }C$ and the temperature of ice $-{15}^{\circ }C$. Find the final temperature of water(in Celcius). (specific heat of ice $=0.5cal/gm{/}^{\circ }C$ and $L=80cal/gm$ ).

Answer 1

As given,
$m_{ice}=2\ast 50gms=100gms$
$m_{water}=250gms$
$(T_i{)}_{ice}=-15$
$(T_i{)}_{water}=-25$
Heat absorbed by ice to get converted from $-{15}^{0}C$ to $0^{0}C$ is $Q_{ice}=100\ast 0.5\ast 15=750cal$
Reduction in temp water $\Delta T=\frac{750}{250\ast s_{water}}=\frac{750}{250\ast 1}=3$
Thus temp of water will decrease to $25-3={22}^{0}C$
To
convert 100 gms of ice at $0^{0}C$ to water at $0^{0}C$, amount
of heat required is $100\ast 80=8000cal$ where $L=80calg{m}^1$
is used.
To convert 250 gm of water at ${22}^{0}C$ to water at $0^{0}C$, the amount of heat released will be $250\ast 1\ast 22=5500cal$
Hence, total heat released is less than the heat required to convert ice $0^{0}C$ to water at $0^{0}C$.
$\therefore aportionoftheicewillgetconvertedfromiceat$0^0 \:C$towaterat$0^0 \:C$.Theamounticeconvertedfromice$0^0 \:C$towaterat$0^0 \:C is $\frac{5500}{80}=\frac{550}{8}gms$
Amount of ice remaining at $0^{0}C$ will be $(100-\frac{550}{8})=\frac{250}{8}=\frac{125}{4}gms$
Amount of water at $0^{0}C$ will be $(100-\frac{125}{4}+250)=\frac{1275}{4}gms$
Final temp of the mixture is $0^{0}C$