Hailstone at 0°C falls from a height of 1 km on an insulating surface converting whole of its kinetic energy into heat. What part of it will melt? (g=10 m/s2) (Latent heat of ice =3.34× 105J/kg)

Question

Hailstone at 0°C falls from a height of 1 km on an insulating surface converting whole of its kinetic energy into heat. What part of it will melt? (g=10 m/s2) (Latent heat of ice =3.34× 105J/kg) (A)  (1/33)  (B)  (1/18)  (C)  (3/3)× 10-4 (D) All of it will melt

Tardigrade Admin · Accepted Answer

Hailstone possesses potential energy due to its height. Let k be the fraction of ice that melts. Hence. potential energy =k × latent heat mgh=kmL ⇒ k=(g h/L)=(10 × 1000/3.34 × 105) ≈ (1/33)

Q. Hailstone at $0^{\circ} C$ falls from a height of $1 km$ on an insulating surface converting whole of its kinetic energy into heat. What part of it will melt? $(g = 10 m / s^{2})$ (Latent heat of ice $= 3.34 \times 1 0^{5} J / k g$ )

$\frac{1}{33}$

$\frac{1}{18}$

$\frac{3}{3} \times 1 0^{- 4}$

All of it will melt

Hailstone possesses potential energy due to its height.
Let $k$ be the fraction of ice that melts.
Hence. potential energy $= k \times$ latent heat $m g h = km L$
$\Rightarrow k = \frac{g h}{L} = \frac{10 \times 1000}{3.34 \times 1 0 ^{5}} \approx \frac{1}{33}$

Q. Hailstone at 0∘C falls from a height of 1km on an insulating surface converting whole of its kinetic energy into heat. What part of it will melt? (g=10m/s2) (Latent heat of ice =3.34×105J/kg)

331​

181​

33​×10−4