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^(-2)`)

Options

A. 8 K
B. 16 K
C. 12 K
D. 24 K

Answer

B

Explanation

Given, mass of hammer, `m = 200 kg,`
steel block of the mass = 200g = 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 = 1/2mv^(2)`
Putting the given values, we get
`1/2 xx 200 xx 8^(2) = 6400 J`
Hence, the `23%` of this energy is converted to heat.
`rArr H = (6400xx23)/(100) = 1472 J`
The rise in temperature of steel,
`Delta T = H/(ms) = (1472)/(460xx0.2) = 16 K`
Hence, the rise in temperature is 16 K.