Question 724: Thermodynamics & KTG

Question

A lead bullet moving with velocity ' $v$ ' strikes a wall and stops. If $50 \%$ of its energy is converted into heat, then the increase in temperature is ( $s=$ specific heat of lead)

Accepted Answer

To determine the increase in temperature when a lead bullet moving with velocity $ v $ strikes a wall and stops, given that 50% of its energy is converted into heat, follow these calculations:
Initial Kinetic Energy (K.E.) of the Bullet
:
$    	ext{K.E.} = \frac{1}{2} mv^2    $
Energy Converted to Heat
:
$    	ext{Energy converted to heat} = 50\% 	ext{ of K.E.} = \frac{1}{2} 	imes \frac{1}{2} mv^2 = \frac{1}{4} mv^2    $
Heat Absorbed by the Bullet
:
Let $ \Delta T $ be the increase in temperature, $ s $ the specific heat capacity of lead, and $ m $ the mass. Then the heat absorbed $ Q $ is:
$    Q = ms \Delta T    $
Equating Mechanical Energy to Thermal Energy
:
Using the formula for the mechanical equivalent of heat $ J $:
$    J = \frac{	ext{Work}}{	ext{Heat}} = \frac{\frac{1}{4} mv^2}{ms \Delta T}    $
Solving for Increase in Temperature
:
Rearrange to find $ \Delta T $:
$    \Delta T = \frac{v^2}{4Js}    $
Thus, the increase in temperature $\Delta T$ of the bullet is expressed as $ \frac{v^2}{4Js} $.

Answer

$\frac{\mathrm{v}^2 \mathrm{~s}}{2 \mathrm{~J}}$

Answer

$\frac{v^2}{4 \mathrm{Js}}$

Answer

$\frac{\mathrm{v}^2 \mathrm{~s}}{\mathrm{~J}}$

Answer

$\frac{2 v^2}{\mathrm{Js}}$

Thermodynamics & KTG

A lead bullet moving with velocity ' v ' strikes a wall and stops. If 50 \% of its energy is converted into heat, then the increase in temperature is ( s= specific heat of lead)

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