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 (specific heat of lead $= 150J/kg,K$)

Question

Dr. Abubacker · Accepted Answer

(c) Since specific heat of lead is given in $Joules,$ hence use $W = Q$ instead of $W = JQ.$
$\Rightarrow \frac{1}{2} 	imes \left( \frac{1}{2}mv^{2} ight) = m.c.\mathrm{\Delta}	heta \Rightarrow \mathrm{\Delta}	heta = \frac{v^{2}}{4c} = \frac{(300)^{2}}{4 	imes 150} = 150℃$

Q. 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 (specific heat of lead $= 150J/kg,K$)

Explanation

Reviewed by Dr. Abubacker

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