A 9.0 kg iron ball is dropped onto a pavement from a height of 110 m.If half of the heat generated goes into warming the ball, find the temperature increase of the ball. (In SI units, the specific heat capacity of iron is 450 J/kg*degree C.)

Answer 1

Set 1/2(M*g*H) equal to M*C*deltaT
amd solve for deltaT, the temperature change of the iron. Note that the mass M cancels out. The specific heat (C) has been provided to you.
H = 110 m
g = 9.81 m/s^2

deltaT = g*H/(2*C)