A hammer of mass $200 \mathrm{~kg}$ strikes a steel block of mass $200 \mathrm{~g}$ with a velocity $8…
A hammer of mass $200 \mathrm{~kg}$ strikes a steel block of mass $200 \mathrm{~g}$ with a velocity $8 \mathrm{~ms}^{-1}$. 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, $\left.=460 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)$
8 K
16 K
12 K
24 K
Solution
Given, mass of hammer, $m=200 \mathrm{~kg}$,
steel block of the mass $=200 \mathrm{~g}=0.2 \mathrm{~kg}$ and specific
heat capacity of steel, $s=460 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
Velocity of hammer, $v=8 \mathrm{~ms}^{-1}$
As we know that,
Kinetic energy, $\mathrm{KE}=\frac{1}{2} m v^2$
Putting the given values, we get
$
=\frac{1}{2} \times 200 \times 8^2=6400 \mathrm{~J}
$
Hence, the $23 \%$ of this energy is converted to heat.
$
\Rightarrow \quad H=\frac{6400 \times 23}{100}=1472 \mathrm{~J}
$
The rise in temperature of steel,
$
\Delta T=\frac{H}{m s}=\frac{1472}{460 \times 0.2}=16 \mathrm{~K}
$
Hence, the rise in temperature is $16 \mathrm{~K}$