First slide

Calorimetry

Question

A lead bullet at 27^oC 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 (M.P. of lead = 327^oC, specific heat of lead :0.03 cal/gm^oC, latent heat of fusion of lead - 6 cal/gm and J = 4.2J/cal)

Moderate

A

410 m/s
B

1230 m/s
C

307.5 m/s
D

None of the above

Solution

If mass of the bullet is m gm, then total heat required for bullet to just melt down
$\begin{array}{l} Q_{1} = mcΔθ + mL = m \times 0 .03 (327 - 27) + m \times 6 \\ = 15 mcal = (15 m \times 4 .2) J \end{array}$
Now when bullet is stopped by the obstacle, the loss in its mechanical energy $= \frac{1}{2} (m \times 10^{- 3}) v^{2} J$
(As m gm $= m \times 10^{- 3}$ kg)
As 25% of this energy is absorbed by the obstacle,
The energy absorbed by the bullet
$Q_{2} = \frac{75}{100} \times \frac{1}{2} {mv}^{2} \times 10^{- 3} = \frac{3}{8} {mv}^{2} \times 10^{- 3} J$
Now the bullet will melt if $Q_{2} \geq Q_{1}$
$i.e. \frac{3}{8} m v^{2} \times 10^{- 3} \geq 15 m \times 4 .2 \Rightarrow v_{min} = 410 m / s$

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