Hailstone at 0?C falls a 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=10m/s2)

Question

Hailstone at 0?C falls a 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=10m/s2)  (A)  (1/33)  (B)  (1/8)  (C)  (1/3)× 10-4  (D) All of it will melt

Tardigrade Admin · Accepted Answer

mgh=kmL k=(gh/L) =(10× 1000/3.36× 105) =(1/33) ( ∵ Latent heat of ice =3.36× 105 text J/kg )

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

$\frac{1}{33}$

$\frac{1}{8}$

$\frac{1}{3} \times 10 - 4$

All of it will melt

$m g h = km L$ $k = \frac{g h}{L}$ $= \frac{10 \times 1000}{3.36 \times 10 ^{5}}$ $= \frac{1}{33}$ ( $∵$ Latent heat of ice $= 3.36 \times 10^{5} J / k g$ )

Q. Hailstone at 0?C falls a 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=10m/s2)

331​

81​

31​×10−4