Question

If a ball of $80\, kg$ mass hits an ice cube and temperature of ball is $100\,^{\circ}C$, then how much ice converted into water ? (Specific heat of ball is $0.2 \,cal \,g^{-1}$, Latent heat of ice $= 80\, cal \,g^{-1}$)

• A. $20\,g$
• B. $200\,g$
• C. $2 \times 10^3\,g$
• D. $2 \times 10^4\,g$

Answer 1

Answer: (D)

$2 \times 10^4\,g$

Answer 2

If $m$ is the mass of ice melted, then heat spent in melting $=$ heat supplied by the ball $mL =sM\Delta T$ $m \times 80 = 0.2 \times (80 \times 1000) \times 100$ or $m =2 \times 10^4\,g$