A lead ball at `25^@`C is dropped from a height of 2 km. It is heated due to air resistance and it is assumed that all of its kinetic energy is used in increasing the temperature of ball. Find the final temperature of the ball. s (ball) = 126 J/kg-K

Explanation

If there is no air resistance, velocity of ball on reaching ground is
`V=sqrt(2gh)=sqrt(2xx10xx2000)=200m//s`
Kinetic energy of ball is
`K=1/2mv^(2)`
`=1/2xxmxx(200)^(2)=2xx10^(4)xxm` Joule
[If mass m is in kg]
If all the kinetic energy is absorbed by the ball as heat and ue to this temperature of ball increased of ball increased by `DeltaT`, we have
`K=msDeltaT`
or `2xx10^(4)xxm=mxx126xxDeltaT`
or `DeltaT=(2xx10^(4))/126=158.73.^(@)C`