A lead bullet at 27,^oC just melts when stopp | vedclass

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(A) Let the mass of the bullet be $m\,g$.The total heat required for the bullet to melt is $Q = mc\Delta	heta + mL$.Given $c = 0.03\,cal/g\,^oC$,$\De

Vedclass · Accepted Answer

$410$

Vedclass · Answer

(A) Let the mass of the bullet be $m\,g$.The total heat required for the bullet to melt is $Q = mc\Delta	heta + mL$.Given $c = 0.03\,cal/g\,^oC$,$\Delta	heta = (327 - 27) = 300\,^oC$,and $L = 6\,cal/g$.$Q = m(0.03 	imes 300 + 6) = m(9 + 6) = 15m\,cal$.Converting to Joules: $Q = 15m 	imes 4.2 = 63m\,J$.The kinetic energy of the bullet is $K = \frac{1}{2}mv^2$ (where $m$ is in $kg$).Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet.$0.75 	imes (\frac{1}{2}mv^2) = 63m$ (where $m$ is in $g$,so $m_{kg} = m 	imes 10^{-3} kg$).$0.75 	imes \frac{1}{2} 	imes (m 	imes 10^{-3}) 	imes v^2 = 63m$.$0.375 	imes 10^{-3} 	imes v^2 = 63$.$v^2 = \frac{63}{0.375 	imes 10^{-3}} = \frac{63000}{0.375} = 168000$.$v = \sqrt{168000} \approx 409.87\,m/s \approx 410\,m/s$.

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