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Q.
$100\, g$ of ice at $0^{\circ} C$ is mixed with $100\, g$ of water $80^{\circ} C$. The final temperature of the mixture will be :
Thermal Properties of Matter
Solution:
Heat given out by $100\, g$ water at $80^{\circ} C$ when it cools to $0^{\circ} C$
$H_{1}=100 \times 1 \times(80-0)=8000\, J$
Heat required by $100\, g$ ice at $0^{\circ} C$ to melt,
$H_{2}=100 \times 80=8000\, J$
As $H_{1}=H_{2}$ so, all of the ice will melt and temperature of mixture is $0^{\circ} C$.