Question

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 = $460 \; J \; kg^{-1} \; K^{-1}$)

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

Answer 1

Answer: (B)

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, $KE =\frac{1}{2} m v^{2}$
Putting the given values, we get
$=\frac{1}{2} \times 200 \times 8^{2}=6400\, J$
Hence, the $23 \%$ of this energy is converted to heat.
$\Rightarrow H=\frac{6400 \times 23}{100}=1472\, J$
The rise in temperature of steel,
$\Delta T=\frac{H}{m s}=\frac{1472}{460 \times 0.2}=16\, K$
Hence, the rise in temperature is $16\, K$.