Question

$1$ g of ice is mixed with $1\, g$ of steam. At thermal equilibrium, the temperature of the mixture is

• A. $0^{\circ}C$
• B. $100^{\circ}C$
• C. $50^{\circ}C$
• D. $55^{\circ}C$

Answer 1

Answer: (B)

$100^{\circ}C$

Answer 2

Key concept According to principle of calorimetry states that total heat given by a hotter body is equal to the total heat received by colder body
i.e. heat lost by hotter body = heat gained by colder body
Heat required to melt $1 g$ of ice at $0^{\circ} C$ into $1 g$ of water at $0^{\circ} C$ is $80 \,cal$.
Heat required to convert $1 \,g$ of water at $0^{\circ} C$ into $1 g$ of water at $100^{\circ} C =1 \times 1 \times 100=100 \,cal$
Heat required to condense $1 g$ of steam
$=1 \times 540 cal =540 \,cal$
Clearly, whole of steam is not condensed. So, temperature of the mixture is $100^{\circ} C$.