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 $=\text{ }0.120\text{ }J{{g}^{-1}}{{C}^{-1}}$ )

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

Answer 1

Answer: (C)

$90{}^\circ C$

Answer 2

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