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  4. One kilogram of ice at 0° C is mixed with one kilogram of water at 80° C. The final temperature of the mixture is (Take specific heat of water =4200 kJ / kg - ° C, Latent heat of ice =336 kJ / kg )

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Q. One kilogram of ice at $0^{\circ} C$ is mixed with one kilogram of water at $80^{\circ} C$. The final temperature of the mixture is (Take specific heat of water $=4200\, kJ / kg -{ }^{\circ} C$, Latent heat of ice $=336\, kJ / kg$ )

BITSATBITSAT 2010

Solution:

It is known that For water and ice mixing
$\theta_{mix}=\frac{m_{W} \theta_{W}-\frac{m_{i} L_{i}}{C_{W}}}{m_{i}+m_{W}}$
$\theta_{\operatorname{mix}}=\frac{m_{W} \theta_{W}-\frac{m_{i} L_{i}}{c_{W}}}{m_{i}+m_{W}}$
$\because m _{ i }= m _{ w }$
$\Rightarrow \theta_{mix}=\frac{\theta_{ W }-\frac{ L _{ i }}{ CW }}{2}$
$=\frac{80-\frac{336}{4.2}}{2}=0^{\circ} C$