1. Tardigrade
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  3. Physics
  4. A lead bullet at 27° C just melts when stopped by an obstacle. Assuming that 25 % of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking is nearly (Melting point of lead =327° C, specific heat of lead =0.03 cal / g ° C, latent heat of fusion of lead =6 cal / g and .J=4.2 J cal -1)

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Q. A lead bullet at $27^{\circ} C$ just melts when stopped by an obstacle. Assuming that $25 \%$ of heat is absorbed by the obstacle, then the velocity of the bullet at the time of striking is nearly (Melting point of lead $=327^{\circ} C$, specific heat of lead $=0.03 \,cal / g ^{\circ} C$, latent heat of fusion of lead $=6 cal / g$ and $\left.J=4.2 \,J \,cal ^{-1}\right)$

Thermodynamics

Solution:

$(75 \%) \frac{1}{2} m v^{2}=(m s \Delta T+m L)$
$\left(\frac{3}{4}\right) \frac{m v^{2}}{2}=m\left(0.03 \times 4.2 J \times 10^{3} kg ^{-1} \times{ }^{\circ} C ^{-1}\right)$
$(327-27)^{\circ} C +m\left(6 \times 4.2 J \times 10^{3} kg ^{-1}\right)$
$\frac{3}{8}\left(v^{2}\right)=(37.8+25.2) \times 10^{3}$
$v=\sqrt{\frac{8 \times 63000}{3}} \approx 410 m s ^{-1}$