Thermodynamics

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A lead ball at 30°C is dropped from a height of 6.2 km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. The molten substance falls slowly on the ground. If the specific heat of lead = 126 Jkg^-1 ^oC^-1 and melting point of lead= 130°C and suppose that any mechanical energy lost is used to heat the ball, then the latent heat of fusion of lead is

2.4 × 10⁴ J Kg^-1

3.6 × 10⁴ J Kg^-1

7.6 × 10² J kg^-1

4.94 × 10³ J kg^-1

Solution

4.94 × 10³ J kg^-1

Given:-

T_m = 130^o C

h = 6.2 km = 6200 m

S = 126 J kg^-1 ^oC^-1

Let the mass of the lead ball be m and its latent heat of fusion be L.

The initial temperature of the lead ball T₁ = 30^o C

The potential energy of the ball gets converted into heat melts the ball.

∴ mgh = mS ( T_m - T₁ ) + mL

⇒ m (10) ( 6200 ) = m (126 ) (130 - 30 ) + mL

⇒ 6200 = 12600 + L

⇒ L = 4.94 × 10⁴ J Kg^-1

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Thermodynamics

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