Question

$500 \,g$ of water and $100\, g$ of ice at $0^{\circ}C$ are in a calorimeter whose water equivalent is $40\, g$. $10\, g$ of steam at $100^\circ C$ is added to it. Then water in the calorimeter is : (Latent heat of ice $= 80\, cal/g$, Latent heat of steam $=540\, cal/g$)

• A. $580\,g$
• B. $590\,g$
• C. $600\,g$
• D. $610\,g$

Answer 1

Answer: (B)

$590\,g$

Answer 2

The correct answer is (B) : 590 g
As $1\,g$ of steam at $100^\circ C$ melts $8g$ of ice at $0^\circ C.$
$10\,g$ of steam will melt $8\times10\, g$ of ice at $0^?C$
Water in calorimeter $= 500 + 80 + 10g = 590g$