Question

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

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

Answer 1

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

From ${20}^{o}C$ ice to $0^{o}C$ ice,

$\Delta Q=m{s}_{ice}\Delta T$
$=1\times \frac{1}{2}\times 20=10kcal(s_{ice}\approx 0.5kcal/k{g}^{o}C)$

From $0^{o}C$ ice to $0^{o}C$ water,

$\Delta Q=m{L}_f$
$=1\times 80=80kcal(L_f=80kcal/k{g}^{o}C)$

From $0^{o}C$ water to ${100}^{o}C$ water,

$\Delta Q=m{s}_{water}\Delta T$
$=1\times 1\times 100=100kcal(s_{water}=1kcal/kg)$

Total heat required $=10+80+100=190kcal$

Heat lost by steam (${200}^{o}C$ to ${100}^{o}C$ steam)
$\Delta Q=m{s}_{steam}\Delta T$
$=1\times 540=540kcal(s_{steam}\approx 0.5kcal/k{g}^{o}C)$

From ${100}^{o}C$ steam to ${100}^{o}C$ water,
$\Delta Q=m{L}_{v}T$
$=1\times \frac{1}{2}\times 100=50kcal(L_v=540kcal/kg)$

Total heat lost by steam $=50+540=590kcal$

From steam at ${100}^{o}C$, only $140kcal(=190-50)$ is required. For that, mass of steam that needs to be converted to water is

$m=\frac{140kcal}{540kcal/kg}=\frac{7}{27}kg$

So final amount of water at ${100}^{o}C$ $=1+\frac{7}{27}=\frac{34}{27}kg$

Final amount of steam at ${100}^{o}C$$=1-\frac{7}{27}=\frac{20}{27}kg$