A lead bullet of mass 10 g travelling at 300 | vedclass

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(C) Given: Mass $m = 10 \ g = 0.01 \ kg$, velocity $v = 300 \ m/s$, specific heat $s = 150 \ J/kg \cdot ^{\circ}C$.Kinetic energy of the bullet $KE =

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$150$

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(C) Given: Mass $m = 10 \ g = 0.01 \ kg$, velocity $v = 300 \ m/s$, specific heat $s = 150 \ J/kg \cdot ^{\circ}C$.Kinetic energy of the bullet $KE = \frac{1}{2}mv^2 = \frac{1}{2} 	imes 0.01 	imes (300)^2 = 0.005 	imes 90000 = 450 \ J$.Heat absorbed by the bullet $Q = 50\% 	ext{ of } KE = 0.5 	imes 450 = 225 \ J$.Using the formula $Q = ms\Delta T$, where $\Delta T$ is the change in temperature:$225 = 0.01 	imes 150 	imes \Delta T$$225 = 1.5 	imes \Delta T$$\Delta T = \frac{225}{1.5} = 150^{\circ}C$.Thus, the increase in temperature is $150^{\circ}C$.

$A$ lead bullet of mass $10 \ g$ travelling at $300 \ m/s$ strikes against a block of wood and comes to rest. Assuming $50\%$ of heat is absorbed by the bullet, the increase in its temperature is (Specific heat of lead $= 150 \ J/kg \cdot ^{\circ}C$) (in $^{\circ}C$)

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