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)

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 × 103 g (D) 2 × 104 g

Tardigrade Admin · Accepted Answer

If m is the mass of ice melted, then heat spent in melting = heat supplied by the ball mL =sMΔ T m × 80 = 0.2 × (80 × 1000) × 100 or m =2 × 104 g

Q. If a ball of $80 k g$ 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 $0.2 c a l g^{- 1}$ , Latent heat of ice $= 80 c a l g^{- 1}$ )

$20 g$

$200 g$

$2 \times 1 0^{3} g$

$2 \times 1 0^{4} g$

If $m$ is the mass of ice melted, then heat spent in melting $=$ heat supplied by the ball
$m L = s M Δ T$
$m \times 80 = 0.2 \times (80 \times 1000) \times 100$
or $m = 2 \times 1 0^{4} g$

Q. If a ball of 80kg mass hits an ice cube and temperature of ball is 100∘C, then how much ice converted into water ? (Specific heat of ball is 0.2calg−1, Latent heat of ice =80calg−1)

20g

200g

2×103g

2×104g