A body of specific heat capacity 450J kg-1 K-1 falls to the ground from rest through a vertical height of 20m. Assuming conservation of energy, calculate th...

A body of specific heat capacity 450J kg^-1 K^-1 falls to the ground from rest through a vertical height of 20m. Assuming conservation of energy, calculate the change in temperature of the body on striking the ground level. (g = 10ms^-2)

Answer Details

When the body falls, its potential energy is converted to kinetic energy, which is then dissipated on striking the ground as heat. The amount of heat produced is equal to the initial potential energy of the body. Therefore, we can equate the potential energy to the heat energy produced by the body. Potential energy of the body = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height fallen. Kinetic energy of the body just before striking the ground = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height fallen. Heat produced on striking the ground = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height fallen. Heat produced = Change in temperature x Mass of the body x Specific heat capacity of the body We can equate the two expressions for heat produced to obtain: mgh = ΔT x m x c where c is the specific heat capacity of the body. Simplifying, we obtain: ΔT = gh/c Substituting the given values, we obtain: ΔT = (10 x 20)/450 = 4/9^oC Therefore, the change in temperature of the body on striking the ground is 4/9^oC. Answer: Option (c).