Question

A lead bullet (specific beat = $0.032cal/g^{\circ }C$) is completely stopped when it strikes a target with a velocity of $300m/s$. The heat generated is equally shared by the bullet and the target.The rise in temperature of bullet will be.

• A. ${16.7}^{o}C$
• B. ${1.67}^{o}C$
• C. ${167.4}^{o}C$
• D. $267.4^{o}C$

Answer 1

The correct option is C ${167.4}^{o}C$

Given that,

Rise temperature $=\theta$

Mass of the bullet $=m$

Now, the kinetic energy associated with the bullet

$K.E=\frac{1}{2}m{v}^2$

$K.E=\frac{1}{2}\times m\times {\left(300\right)}^2$

$K.E=m\times 45000J$

Now, according to law of conservation of energy

$\frac{m\times 45000}{2}=m\times 4.2\times {10}^3\times 0.032\times \theta$

$\theta =\frac{45000}{2\times 4.2\times {10}^3\times 0.032}$

$\theta =\frac{450000}{2\times 42\times 32}$

$\theta ={167.4}^{0}C$

Hence, the rise temperature of bullet is ${167.4}^{0}C$