A hammer of mass 200 ~kg strikes a steel block of mass 200 ~g with a velocity 8 ~ms ⁻¹ . If 23 % of the energy is utilized to heat the steel block, the rise in temperature of the block is (specific heat capacity of steel, .=460 ~J ~kg ⁻¹ ~K ⁻¹ )

Step-by-step solution

Given, mass of hammer, m=200 ~kg , steel block of the mass =200 ~g =0.2 ~kg and specific heat capacity of steel, s=460 ~J ~kg ⁻¹ ~K ⁻¹ Velocity of hammer, v=8 ~ms ⁻¹ As we know that, Kinetic energy, KE = 1 2 m v^2 Putting the given values, we get = 1 2 200 8^2=6400 ~J Hence, the 23 % of this energy is converted to heat. H= 6400 23 100 =1472 ~J The rise in temperature of steel, T= H m s = 1472 460 0.2 =16 ~K Hence, the rise in temperature is 16 ~K .