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Q.
$1\, kg$ ice at $-20^{\circ} C$ is mixed with $1\, kg$ steam at $200^{\circ} C$. The equilibrium temperature and mixture content is
Thermal Properties of Matter
Solution:
Let equilibrium temperature is $100^{\circ} C$ heat required to convert $1\, kg$ ice at $-20^{\circ} C$ to $1 kg$ water at $100^{\circ} C$ is equal to
$H_{1}=1 \times \frac{1}{2} \times 20+1 \times 80+1 \times 1 \times 100=190\, Kcal$
heat release by steam to convert $1\, kg$ steam at $200^{\circ} C$
to $1\, kg$ water at $100^{\circ} C$ is equal to
$H_{2}=1 \times \frac{1}{2} \times 100+1 \times 540=590\, Kcal$
$1\, kg$ ice at $-20^{\circ} C$
$=H_{1}+1\, kg$ water at $100^{\circ} C\,\, .....(1)$
$1\, kg$ steam at $200^{\circ} C$
$=H_{2}+1\, kg$ water at $100^{\circ} C\,\, .....(2)$
by adding equation $(1)$ and $(2)$
$1\, kg$ ice at $-20^{\circ} C +1\, kg$ steam at $200^{\circ} C =H_{1}+H_{2}+$
$2\, kg$ water at $100^{\circ} C$.
Here heat required to ice is less than heat supplied by steam so mixture equilibrium temperature is $100^{\circ} C$ then steam is not completely converted into water.
So mixture has water and steam which is possible only at $100^{\circ} C$ mass of steam which converted into water is equal to
$m=\frac{190-1 \times \frac{1}{2} \times 100}{540}=\frac{7}{27} kg$
In mixture content
mass of steam $=1-\frac{7}{27}=\frac{20}{27} kg$
mass of water $=1+\frac{7}{27}=\frac{34}{27} kg$