Question

A $2\text{g}$ bullet moving with a speed of $210\text{m/s}$ brought to a sudden stoppage by an obstacle. The total heat produced goes to the bullet. If the specific heat of the bullet is $0.03{\text{cal/g}}^{\circ }\text{C}$ the rise in its temperature will be

• A. ${175}^{\circ }\text{C}$
• B. ${180}^{\circ }\text{C}$
• C. ${150}^{\circ }\text{C}$
• D. ${125}^{\circ }\text{C}$

Answer 1

The correct option is A ${175}^{\circ }\text{C}$

Mass of bullet; $m=2\text{g}$
Initial velocity of bullet, $v=210\text{m/s}$
$\therefore$ Initial kinetic energy of bullet $=\frac{1}{2}m{v}^2$
Specific heat of bullet $c=0.03{\text{cal/g}}^{\circ }=0.03\times 4200{\text{J/kg}}^{\circ }\text{C}$
According to the question, kinetic energy of bullet gets converted into heat and increases the temperature $\left(\Delta T\right)$ of bullet.
$\therefore \frac{1}{2}m{v}^2=\left(mc\Delta T\right)$
$\Rightarrow \frac{1}{2}\times 2\times {10}^{-3}\times (210{)}^2=2\times {10}^{-3}\times 0.03\times 4200\times \Delta T$
$\Rightarrow \Delta T={175}^{\circ }\text{C}$