Question

If a ball of 80 kg mass hits an ice cube and temperature of ball is 100°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. $2 \times 10^{3}\text{ g}$
• D. $2 \times 10^{4}\text{ g}$

Answer 1

Answer: (D)

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

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 × 80 = 0.2×(80×1000)×100 or $m = 2 \times 10^{4}g$