Question

$1g$ of steam is sent into $1g$ of ice. At thermal equilibrium, the resultant temperature of mixture is

• A. ${270}^{\circ }C$
• B. ${230}^{\circ }C$
• C. ${100}^{\circ }C$
• D. ${120}^{\circ }C$

Answer 1

Answer: (C)

${100}^{\circ }C$

Latent heat of vaporisation = $540cal/g$.

Latent heat of fusion = $80cal/g$.

Specific heat of water = $1cal/g°C$.

So, heat required to convert ice to steam = $80+100=180cal$.

Steam can provide more than the required energy. So, all the ice will be converted to steam plus water mixture at 100 degree celsius. Thus, the final mixture will be that of steam and water in equilibrium with each other at ${100}^{0}C$.