Question

An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inelastic. Assuming all the ball's kinetic energy goes into heating the ball, find its change in temperature.

Answer 1

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = $87.6*10^{-3}*9.81*7$

When leads gets heated by a temperature ΔT energy needed = mcΔT

                                                                      = $87.6*10^{-3}*0.128*10^3$ΔT

Comparing both the equations

                      $87.6*10^{-3}*9.81*7$ = $87.6*10^{-3}*0.128*10^3$ΔT

                        ΔT = 0.536 K

                        Change in temperature same in degree and kelvin scale

                                      So ΔT = 0.536 $^0C$