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10-1.Thermometry, Thermal Expansion and Calorimetry
medium
A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the whole mass melts be $m$, specific heat $S,$ initial temperature $25°C$ melting point $ 475°C$ and the latent heat $L.$ Then $v$ is given by
A
$mL = mS\,(475 - 25) + \frac{1}{2} \cdot \frac{{m{v^2}}}{J}$
B
$mS(475 - 25) + mL = \frac{{m{v^2}}}{{2J}}$
C
$mS\,(475 - 25) + mL = \frac{{m{v^2}}}{J}$
D
$mS\,(475 - 25) - mL = \frac{{m{v^2}}}{{2J}}$
Solution
(b) Firstly the temperature of bullet rises up to melting point, then it melts. Hence according to $W = JQ$.
$\Rightarrow$ $\frac{1}{2}m{v^2} = J.[m.c.\Delta \theta + mL] = J[m\,S\,(475 – 25) + mL]$
$\Rightarrow$$mS(475 – 25) + mL = \frac{{m{v^2}}}{{2J}}$
Standard 11
Physics
