A lead bullet at 27,^{circ}C just melts when | vedclass

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Question

(A) Let the mass of the bullet be $m\,g$.The heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta	heta + mL$.

Vedclass · Accepted Answer

$410$

Vedclass · Answer

(A) Let the mass of the bullet be $m\,g$.The heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta	heta + mL$.Substituting the values: $Q_1 = m 	imes 0.03 	imes (327 - 27) + m 	imes 6 = m 	imes 0.03 	imes 300 + 6m = 9m + 6m = 15m\,cal$.Converting this to Joules: $Q_1 = 15m 	imes 4.2 = 63m\,J$.The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$,where $m$ is in $kg$. Since $m$ is in $g$,$m_{kg} = m 	imes 10^{-3}$.So,$K.E. = \frac{1}{2} 	imes (m 	imes 10^{-3}) 	imes v^2 = 0.5 	imes 10^{-3} mv^2\,J$.Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet to melt it.Therefore,$0.75 	imes (0.5 	imes 10^{-3} mv^2) = 63m$.$0.375 	imes 10^{-3} v^2 = 63$.$v^2 = \frac{63}{0.375 	imes 10^{-3}} = \frac{63}{0.000375} = 168000$.$v = \sqrt{168000} \approx 409.88\,m/s \approx 410\,m/s$.

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