A steel ball of mass 0.1 kg falls from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, find the rise in temperature of the ball ? (specific heat of steel = 460 JKg^(-1)K^(+) , g = 10 ms^(-2))

A steel ball of mass 0.1 kg falls from a height of 10 m and bounces to

1.	A steel ball of mass 0.1 kg falls from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, find the rise in temperature of the ball ? (specific heat of steel = 460 JKg^(-1)K^(+) , g = 10 ms^(-2))
Answer» Solution :Loss in gravitational PE = HEAT energy , `0.1 XX 10(10 - 5.4) = 0.1 xx 460 xx Delta theta` `IMPLIES Deltatheta = 0.1^(0)C` (rise in temperature)