Question 665: Thermodynamics & KTG

Question

A stationary object at $4^{\circ} \mathrm{C}$ and weighing 3.5 kg falls from a height of 2000 m on snow mountain at $0^{\circ} \mathrm{C}$. If the temperature of the object just before hitting the snow is $0^{\circ} \mathrm{C}$ and the object comes to rest immediately then the quantity of ice that melts is (Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$, Latent heat of ice $=3.5 	imes 10^5 \mathrm{~J} / \mathrm{kg}$ )

Accepted Answer

Step 1: Calculate potential energy (P.E.)
The object has potential energy because it is 2000 m above the ground. We use the formula:
$$ 	ext{P.E.} = mgh $$
Here,

$m = 3.5\, \mathrm{kg}$ (mass of object),

$g = 10\, \mathrm{m/s}^2$ (acceleration due to gravity),

$h = 2000\, \mathrm{m}$ (height).
Step 2: Understand energy conversion
When the object falls, its potential energy turns into heat energy, which is used to melt some of the ice. All the energy goes into melting the ice because the object immediately comes to rest.
We set the energy from falling equal to the heat needed to melt $M$ kg of ice:
$$ 	ext{Latent heat needed} = ML $$
$$ 	ext{Potential energy lost} = mgh $$
So,

$$ ML = mgh $$
Step 3: Solve for the mass of melted ice ($M$)
Given latent heat of ice, $L = 3.5 	imes 10^5\, \mathrm{J/kg}$.
Plug in the values:
$$ M 	imes 3.5 	imes 10^5 = 3.5 	imes 10 	imes 2000 $$
$$ M 	imes 3.5 	imes 10^5 = 70,000 $$
$$ M = \frac{70,000}{3.5 	imes 10^5} $$
$$ M = 0.2\, \mathrm{kg} $$
So, $0.2\, \mathrm{kg}$ or $200$ grams of ice melts.

Answer

2 gram

Answer

20 gram

Answer

200 gram

Answer

2 kg

Thermodynamics & KTG

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