Question

A sealed vaporisationins $100$ g of water and if the water absorbs $100$ kJ of heat. Due to absorption of heat temperature is raised from ${20}^{o}C$. Given that water has the specific heat $4.186$ kJ/kg${}^o$C, a boiling point of ${100}^{o}C$, and a heat of vaporization of $2,260$ kJ/kg. Calculate the final temperature of the water.

• A. ${100}^{o}C$
• B. ${119}^{o}C$
• C. ${143}^{o}C$
• D. ${1830}^{o}C$
• E. ${239}^{o}C$

Answer 1

The correct option is A ${100}^{o}C$

Heat required to raise temperature to ${100}^{\circ }=mc\Delta T$

$=0.1kg\times 4.186kJ/k{g}^{\circ }C\times {80}^{\circ }C$

$33.48kJ<100kJ$

Thus the temperature would rise to ${100}^{\circ }$

Latent heat required to convert water to vapour=$m{L}_{vap}$

$=0.1kg\times 2260kJ/kg$

$=226kJ>(100kJ-33.48kJ)$

Thus the water would not convert to vapour and final temperature would remain ${100}^{\circ }C$