Question

A bullet of mass $5g$, travelling with a speed of $210m/s$, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is $0.03cal/gm°C$ ($1cal=4.2\times {10}^{7}ergs$) close to:

• A. $38.4°C$
• B. $87.5°C$
• C. $83.3°C$
• D. $119.2°C$

Answer 1

The correct option is B

$87.5°C$

Step 1: Given data

Mass of bullet, $m=5g$

Speed of bullet,$v=210m/s$

Specific heat, $Q=0.03cal/gm°C$

Step 2: Find rise of temperature of the bullet

According to question,

$$\begin{gathered} \frac{1}{2}K.E.=Q \\ \frac{1}{4}m{v}^2=mc\Delta T \\ \frac{1}{4}\times v^2=\left[\right[0.03\times 4.2J]/{10}^{-3}]\Delta T \\ {v}^2=4\times 126\Delta T \\ {\left(210\right)}^2=4\times 126\Delta T \\ \Delta T=87.5°C \end{gathered}$$

Hence, option B is correct.