Question

$1kg$ ice at $-20^{\circ }C$ is mixed with $1kg$ steam at $200^{\circ }C$. The equilibrium temperature and mixture content is

• A. $80^{\circ }C$, and mixture content 2 kg
• B. $110^{\circ }C$, and mixture content 2 kg steam
• C. $100^{\circ }C$, and mixture content $\frac{10}{17}kg$ steam and $\frac{24}{17}kg$ water
• D. $100^{\circ }C$, and mixture content $\frac{20}{27}kg$ steam and $\frac{34}{27}kg$ water

Answer 1

Answer: (D)

Let equilibrium temperature is $100^{\circ }C$ heat required to convert $1kg$ ice at $-20^{\circ }C$ to $1kg$ water at $100^{\circ }C$ is equal to
$H_1=1\times \frac{1}{2}\times 20+1\times 80+1\times 1\times 100=190Kcal$
heat release by steam to convert $1kg$ steam at $200^{\circ }C$
to $1kg$ water at $100^{\circ }C$ is equal to
$H_2=1\times \frac{1}{2}\times 100+1\times 540=590Kcal$
$1kg$ ice at $-20^{\circ }C$
$=H_1+1kg$ water at $100^{\circ }C.....(1)$
$1kg$ steam at $200^{\circ }C$
$=H_2+1kg$ water at $100^{\circ }C.....(2)$
by adding equation $(1)$ and $(2)$
$1kg$ ice at $-20^{\circ }C+1kg$ steam at $200^{\circ }C=H_1+H_2+$
$2kg$ 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 $=1+\frac{7}{27}=\frac{34}{27}kg$