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  4. A lead bullet, of initial temperature 27?C and speed v km/h penetrates into a solid object and melts. If 50% of the kinetic energy is used to heat it, the value of v in km/h is (for lead melting point = 600 K, latent heat of fusion =2.5× 104 text J text kg-1; specific heat =125J text kg-1K-1 )

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Q. A lead bullet, of initial temperature 27?C and speed v km/h penetrates into a solid object and melts. If 50% of the kinetic energy is used to heat it, the value of v in km/h is (for lead melting point = 600 K, latent heat of fusion $ =2.5\times 104\text{ }J\text{ }k{{g}^{-1}}; $ specific heat $ =125J\text{ }k{{g}^{-1}}{{K}^{-1}} $ )

EAMCETEAMCET 2008Thermal Properties of Matter

Solution:

Heat energy required to just melt the bullet $ Q={{Q}_{1}}+{{Q}_{2}} $ Here, $ {{Q}_{1}}=ms\Delta \theta $ $ =m\times 125\times (327-27) $ $ =3.75\times {{10}^{4}}m $ $ {{Q}_{2}}=mL=m\times 2.5\times {{10}^{4}} $ $ =2.5\times {{10}^{4}}m $ $ \therefore $ $ Q=6.25\times {{10}^{4}}\,m $ If v be the speed of bullet, then 50% of $ \frac{1}{2}m{{v}^{2}} $ should be equal to Q. Thus, $ 0.5\times \frac{1}{2}m{{v}^{2}}=6.25\times {{10}^{4}}m $ $ \Rightarrow $ $ v=\sqrt{\frac{6.25\times {{10}^{4}}\times 2}{0.5}} $ $ =500\,m/s=\,1800\,km/h $