Question

When a bullet leaves a rifle it has a lot of kinetic energy, most of which gets lost as heat. Imagine that a 9.8kg lead bullet is fired into a ort upon which the actual target was sitting . The bullets initial speed is 760 m/s, and as it comes to rest half of its kinetic energy goes into heating the bullet. How much does the bullets temperature increase

Answer 1

$According \,\,to\,\,the\,\,given\,\,information\\ \frac{1}{2} \times total \,\,kinetic\,\,energy=total\,\,heat\,\,energy\,\,generated\\\\ \frac{1}{2}\times \left (\frac{1}{2}mv^2 \right )=mC_{Pb}\Delta T\\\\ Here\,\,\,\,m=9.8\,\,kg,\,\,\,\,\,\,v=760\,\,m/s\\ C_{Pb}=128\,\, \frac{J}{kg\,\,^{\circ}C}\\ \Delta T=increase\,\,in\,\,temperature$

$\therefore \,\,\,\, \frac{1}{2}\times \left (\frac{1}{2}\times9.8\times 760^2 \right )=9.8\times128\,\,\Delta T\\\\ \Rightarrow \Delta T=\frac{9.8\times760^2}{2\times2\times9.8\times128}\,\,\,\,^{\circ}C\\\\ \Rightarrow \Delta T=1128.125\,\,^{\circ}C$



$Specific\,\, heat\,\,of\,\,leadis\,\,taken\,\,from\,\,internet\,\,as\,\,128\,\,\frac{J}{kg\,\,^{\circ}C}$


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