Question

If a ball of $80kg$ 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 $0.2cal{g}^{-1}$, Latent heat of ice $=80cal{g}^{-1})$

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

Answer 1

Answer: (D)

$2\times {10}^{4}g$

Given: The mass of the ball is $80kg$.

The temperature of the ball is ${100}^{\circ }C$

To find: The mass of the ice converted into water.

If m is the mass of ice melted, then

heat spent in melting = heat supplied by the ball

$mL=Ms\Delta T$

$m\times 80=0.2\times (80\times 1000)\times 100$

or $m=2\times {10}^{4}g$