Question

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

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

Answer 1

The correct option is D ${100}^{\circ }C\text{and mixture content}\frac{20}{27}kg\text{steam and}\frac{34}{27}kg\text{water}$.

Formula used: $Q=mL,Q=ms\Delta \theta$
Given : $m_1=1kg,{\theta }_1=-{20}^{\circ }C,m_2=1kg,{\theta }_2={200}^{\circ }C$

Heat required to convert $1kg\text{ice at}-{20}^{\circ }C\text{to}1kg\text{water at}{100}^{\circ }C\text{is equal to}$
$H_1=1\times \frac{1}{2}\times 20+1\times 80+1\times 100=190kcal$

Heat released by steam to convert $1kg\text{steam at}{200}^{\circ }C\text{to}1kg\text{water at}{100}^{\circ }C\text{is equal to}$
$H_2=1\times \frac{1}{2}\times 100+1\times 540=590kcal$

As we can see ice will melt and will reach boiling temperature, to further convert into steam it would require further $540cal$ energy which is not available as we can see from $H_2$ value.
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 mixer has water and steam which is possible only at ${800}^{\circ }C$.
some part of steam will get converted into water and let that be $m$.
$\text{From}{H}_1\text{and}{H}_2$ values,
$50+mL=190$
$m=\frac{190-50}{540}=\frac{7}{27}kg$
So, mixture content:
Mass of steam $=1-\frac{7}{27}=\frac{20}{27}kg$
Mass of water $=1+\frac{7}{27}=\frac{34}{27}kg$

FINAL ANSWER: (d).