Question

A lead bullet travelling at 210m/s enters a block of wood and is brought to rest. Assuming 75% of heat absorbed by the bullet, calculate the increase in temperature (Specific heat of lead 0.03 X 103 cal/kg${}^{o}C$ and J$=$ 4.2 X ${10}^3$ J/K-Cal) nearly

• A. ${131}^{o}C$
• B. ${100}^{o}C$
• C. ${31}^{o}C$
• D. ${69}^{o}C$

Answer 1

The correct option is A ${131}^{o}C$

The kinetic energy of the travelling bullet is,
$\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(210\right)}^2$
Out of this, 75% is absorbed by the bullet in collision.
Energy that raises the temperature,
= $\frac{3}{4}\times \frac{1}{2}m{\left(210\right)}^2$
This is equal to $mC\Delta T$
On equating the 2 energies,
$\frac{3}{4}\times \frac{1}{2}m{\left(210\right)}^2=mC\Delta T$
$\Delta T={131}^{o}C$