An ice block at 0° C is dropped from height 'h' above the ground. What should be the value of ' h ' so that it melts completely by the time it reaches the bottom assuming the loss of whole gravitational potential energy is used as heat by the ice? [Given: .Lf=80 cal / gm ]

Question

An ice block at 0° C is dropped from height 'h' above the ground. What should be the value of ' h ' so that it melts completely by the time it reaches the bottom assuming the loss of whole gravitational potential energy is used as heat by the ice? [Given: .Lf=80 cal / gm ] (A) 33.6 m (B) 33.6 km (C) 8 m (D) 8 km

Tardigrade Admin · Accepted Answer

Applying energy conservation: m g h=m Lf ⇒ h=(Lf/g)=(80 cal / gm /10 m / s 2) =(80 × 4.2 × 1000 J / kg /10 m / s 2) ((336 × 103/10)) ( N - s 2/ kg ) =33.6 km .

Q. An ice block at $0^{\circ} C$ is dropped from height 'h' above the ground. What should be the value of ' $h$ ' so that it melts completely by the time it reaches the bottom assuming the loss of whole gravitational potential energy is used as heat by the ice? [Given: $L_{f} = 80 c a l / g m]$

33.6 m

33.6 km

8 m

8 km

Applying energy conservation :
$m g h = m L_{f}$
$\Rightarrow h = \frac{L _{f}}{g} = \frac{80 c a l / g m}{10 m / s ^{2}}$
$= \frac{80 \times 4.2 \times 1000 J / k g}{10 m / s ^{2}}$
$(\frac{336 \times 1 0 ^{3}}{10}) \frac{N - s ^{2}}{k g}$
$= 33.6 km .$

Q. An ice block at 0∘C is dropped from height 'h' above the ground. What should be the value of ' h ' so that it melts completely by the time it reaches the bottom assuming the loss of whole gravitational potential energy is used as heat by the ice? [Given: Lf​=80cal/gm]

33.6 m

33.6 km

8 m

8 km

Applying energy conservation : mgh=mLf​ ⇒h=gLf​​=10m/s280cal/gm​ =10m/s280×4.2×1000J/kg​ (10336×103​)kgN−s2​ =33.6km.

Q. An ice block at $0^{\circ} C$ is dropped from height 'h' above the ground. What should be the value of ' $h$ ' so that it melts completely by the time it reaches the bottom assuming the loss of whole gravitational potential energy is used as heat by the ice? [Given: $L_{f} = 80 c a l / g m]$

Applying energy conservation :
$m g h = m L_{f}$
$\Rightarrow h = \frac{L _{f}}{g} = \frac{80 c a l / g m}{10 m / s ^{2}}$
$= \frac{80 \times 4.2 \times 1000 J / k g}{10 m / s ^{2}}$
$(\frac{336 \times 1 0 ^{3}}{10}) \frac{N - s ^{2}}{k g}$
$= 33.6 km .$