Question

In a process 10 g of ice at $-5^{o}C$ is converted into the steam at ${100}^{o}C$. If latent heat of fusion of ice is $80cal{g}^{-1}$, then the amount of heat required to convert 10 g of ice at $0^{o}C$ into 10 g of water at same temperature is :

• A. 400 cal
• B. 800 cal
• C. 1200 cal
• D. 1600 cal

Answer 1

Answer: (B)

800 cal

Heat required to convert 10 g of ice into water at a constant temperature $0^{o}C.$

$Q=mL$

$=\left(10g\right)\left(80cal{g}^{-1}\right)$

$Q=800cal$

Hence (B) is the correct answer.