A gun fires a lead bullet of temperature 300 | vedclass

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(C) The total heat required for the process consists of two parts: the heat required to raise the temperature of the bullet to its melting point and t

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$10$

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(C) The total heat required for the process consists of two parts: the heat required to raise the temperature of the bullet to its melting point and the heat required for the phase change (melting).Let $m$ be the mass of the bullet in $kg$.Heat required to raise the temperature from $300 \ K$ to $600 \ K$ is $Q_1 = ms \Delta T = m 	imes 125 	imes (600 - 300) = m 	imes 125 	imes 300 = 37500m \ J$.Heat required for melting is $Q_2 = mL = m 	imes 2.5 	imes 10^4 = 25000m \ J$.Total heat $Q = Q_1 + Q_2 = 37500m + 25000m = 62500m \ J$.Given $Q = 625 \ J$,we have $62500m = 625$.$m = \frac{625}{62500} = 0.01 \ kg$.Converting to grams,$m = 0.01 	imes 1000 = 10 \ g$.

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