Question

A metal 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^o C^{-1} and melting point of lead = 130^oC and suppose that any mechanical energy lost is used to heat the ball, then the latent heat of fusion of lead is:__________.

Answer 1

Step-by-step explanation:

Given

height
`h=6.2 km`

Initial temperature
`T_1=30^(\circ)C`

Specific heat of lead
`c=126 J/kg-^(\circ)C`

Melting Point of Lead
`T_m=130^(\circ)C`

Here Potential Energy is converted to heat energy to melt the lead ball

Sphere ball will first will be heated to
`130^(\circ)C` then it starts melting

thus

`mgh=mc\Delta T+mL`

where
`L=latent\ heat\ of\ fusion`

`\Delta T=`change in Temperature

`gh=c\Delta T+L`

`9.8* 6.2* 1000=126* (130-30)+L`

`L=48,160\ J/kg`

`L=48.16\ kJ/kg`