A 12-kg iron ball is dropped onto a pavement from a height of 90m. Suppose half of the heat generated goes into warming the ball. What is the temperature increase of the ball in degree C

mgh/2=mcΔT

For iron c=444 J/kg•K
ΔT=gh/2c = 9.8•90/444•2=0.99℃

To calculate the temperature increase of the iron ball, you need to consider the energy gained by the ball while it falls and then use this energy to calculate the temperature increase.

First, let's find the potential energy of the ball when it is at a height of 90m. The potential energy (PE) is given by the formula PE = mgh, where m is the mass of the ball (12 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (90m).

PE = 12 kg x 9.8 m/s^2 x 90m
= 10,440 Joules

Since half of the heat generated goes into warming the ball, we can assume that half of the potential energy will be converted into heat.

Heat energy (Q) = 1/2 x PE
= 1/2 x 10,440 Joules
= 5,220 Joules

Now, we can use the heat energy gained (Q) to calculate the temperature increase (ΔT) using the formula ΔT = Q / (m x c), where c is the specific heat capacity of iron.

The specific heat capacity of iron is approximately 450 J/kg°C.

ΔT = 5,220 Joules / (12 kg x 450 J/kg°C)
= 9.17 °C

Therefore, the temperature increase of the iron ball is approximately 9.17 degrees Celsius.