Work Power and Energy 3 Question 8
8. A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is $27^{\circ} C$.
(Melting point of lead $=327^{\circ} C$, specific heat of lead $=0.03 cal / g-{ }^{\circ} C$, latent heat of fusion of lead $=6 cal / g$, $J=4.2 J / cal)$.
(1981, 3M)
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Answer:
Correct Answer: 8. $409.8 m / s$
Solution:
- Heat energy required to just melt the bullet.
$$ \begin{aligned} Q & =Q _1+Q _2 \\ \text { Here, } Q _1 & =m s \Delta \theta \\ & =\left(m \times 10^{3}\right)(0.03 \times 4.2)(327-27) \\ & =\left(3.78 \times 10^{4} m\right) \\ Q _2 & =m L=\left(m \times 10^{3}\right)(6 \times 4.2) \\ & =\left(2.52 \times 10^{4} m\right) \\ \therefore \quad Q & =\left(6.3 \times 10^{4} m\right) \end{aligned} $$
If $v$ be the speed of bullet, then $75 %$ of $1 / 2 m v^{2}$ should be equal to $Q$. Thus,
$0.75 \times \frac{1}{2} \times m \times v^{2}=6.3 \times 10^{4} m$
$\Rightarrow \quad v=409.8 m / s$
