Question

$1g$ of ice at $0^{o}C$ is converted to steam at ${100}^{o}C$. The amount of heat required will be

• A. $1200cal$
• B. $756cal$
• C. $720cal$
• D. $430cal$

Answer 1

Answer: (C)

$720cal$

We have the latent heat of fusion as $80cal/g$, specific heat of water as 1 cal/g/K and latent heat of vaporization as $540cal/g$.
Thus we get the amount of heat required as-
heat required to melt ice+heat required to raise the temperature of water from $0^{o}C$ to ${100}^{o}C$ + heat required ot vaporize water
=$m{L}_{ice}+mC\Delta T+m{L}_{water}=(1\times 80)+(1\times 1\times 100)+(540\times 1)=720cal$