A lead bullet of 10,g travelling at 300,m/s strikes against a block of wood and comes to rest. Assum

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10-1.Thermometry, Thermal Expansion and Calorimetry

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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 $= 150\,J/kg,\,K$ )

A

$100$

B

$125$

C

$150$

D

$200$

Solution

since specific heat of lead is given in Joules, hence use $W=Q$ instead of $W=J Q$

$\frac{1}{2} \times\left(\frac{1}{2} m v^{2}\right)=m \cdot c \cdot \Delta \theta$

$\Delta \theta=\frac{v^{2}}{4 c}=\frac{(300)^{2}}{4 \times 150}=150^{\circ} \mathrm{C}$

Standard 11

Physics

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