Question

A lead bullet of unknown mass is fired with a speed of 180 ms-1 into a tree in which it stops. Assuming that in this process two-third of heat produced goes into the bullet and one- third into wood. The temperature of the bullet rises by (Specific heat of lead $=0.120J{g}^{-1}{C}^{-1}$ )

• A. $140{}^{\circ }C$
• B. $106{}^{\circ }C$
• C. $90{}^{\circ }C$
• D. $100{}^{\circ }C$

Answer 1

Answer: (C)

Since specific heat of lead is given in joule. Specific heat of lead $=0.120J/g{}^{o}C=120J/kg$ The two-third of head produced goes into the bullet So, $m\times s\times \Delta \theta =\frac{2}{3}\times \frac{1}{2}m{v}^2$ $\Delta \theta =\frac{v^2}{3\times s}$ $=\frac{180\times 180}{3\times 120}=90{}^{o}C$