A lead bullet of 10,g travelling at 300,m/s s | vedclass

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Question

(C) The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$.Given mass $m = 10\,g = 0.01\,kg$,velocity $v = 300\,m/s$,and specific heat $c = 150\

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$150$

Vedclass · Answer

(C) The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$.Given mass $m = 10\,g = 0.01\,kg$,velocity $v = 300\,m/s$,and specific heat $c = 150\,J/kg\cdot K$.The total heat produced is equal to the kinetic energy lost: $Q = \frac{1}{2}mv^2$.Only $50\,\%$ of this heat is absorbed by the bullet,so the heat absorbed is $Q_{abs} = 0.5 	imes \frac{1}{2}mv^2 = \frac{1}{4}mv^2$.The heat absorbed also equals $mc\Delta	heta$,where $\Delta	heta$ is the change in temperature.Equating the two: $mc\Delta	heta = \frac{1}{4}mv^2$.$\Delta	heta = \frac{v^2}{4c} = \frac{(300)^2}{4 	imes 150} = \frac{90000}{600} = 150^{\circ}C$.

List-$I$	List-$II$
$(A)$ Heat capacity of body	$(I)$ $J kg^{-1}$
$(B)$ Specific heat capacity of body	$(II)$ $J K^{-1}$
$(C)$ Latent heat	$(III)$ $J kg^{-1} K^{-1}$
$(D)$ Thermal conductivity	$(IV)$ $J m^{-1} K^{-1} s^{-1}$

$A$ lead bullet of $10\,g$ 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 ........ $^oC$ (Specific heat of lead $= 150\,J/kg\cdot K$)

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