A lead bullet penetrates into a solid object and melts. Assuming that $40 %$ of its kinetic energy is used to heat it, the initial speed of bullet is
(Given, initial temperature of the bullet $= 127 ° C$ ,
Melting point of the bullet $= 327 ° C$ ,
Latent heat of fusion of lead $= 2.5 \times 10^{4} J {kg}^{- 1}$ ,
Specific heat capacity of lead $= 125 J kg K^{- 1}$ )

Question

A lead bullet penetrates into a solid object and melts. Assuming that

40 %

of its kinetic energy is used to heat it, the initial speed of bullet is
(Given, initial temperature of the bullet

= 127 ° C

,
Melting point of the bullet

= 327 ° C

,
Latent heat of fusion of lead

= 2.5 \times 10^{4} J {kg}^{- 1}

,
Specific heat capacity of lead

= 125 J kg K^{- 1}

)

Aicharya · Accepted Answer

Loss of kinetic energy of the lead bullet $=$ Heat generated to raise its temperature from $127 ° C$ to $327 ° C$ .

$40 %$ loss of the kinetic energy of the lead bullet $=$ Heat gained by the lead bullet to raise its temperature.

$0.4 (\frac{1}{2} m v^{2}) = m \times [S_{Pb} \times ∆ T + m \times L_{Pb}]$

$0.4 \times \frac{1}{2} m v^{2} = m S_{Pb} (T_{2} - t_{1}) + m L_{Pb}$

$0.2 m v^{2} = m \times [125 \times (327 - 127) + 2.5 \times 10^{4}]$

$v^{2} = 5 \times [125 \times 200 + 2.5 \times 10^{4}]$

$v^{2} = 5 \times [25000 + 25000]$

$v = 500 m s^{- 1}$

A lead bullet penetrates into a solid object and melts. Assuming that 40 % of its kinetic energy is used to…