A piece of ice (heat capacity = 2100 J kg^{ | vedclass

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(B) The total heat $Q = 420 \ J$ is used for two processes:$1$. Raising the temperature of the entire mass $m$ (in grams) from $-5^{\circ}C$ to $0^{\c

Vedclass · Accepted Answer

$8$

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(B) The total heat $Q = 420 \ J$ is used for two processes:$1$. Raising the temperature of the entire mass $m$ (in grams) from $-5^{\circ}C$ to $0^{\circ}C$.$2$. Melting $1 \ g$ of ice at $0^{\circ}C$.Given:Mass $m$ in grams $= m 	imes 10^{-3} \ kg$.Specific heat of ice $c = 2100 \ J \ kg^{-1} \ ^{\circ}C^{-1}$.Latent heat of fusion $L = 3.36 	imes 10^5 \ J \ kg^{-1}$.Heat required to raise temperature: $Q_1 = m 	imes 10^{-3} 	imes 2100 	imes (0 - (-5)) = m 	imes 2.1 	imes 5 = 10.5m \ J$.Heat required to melt $1 \ g$ of ice: $Q_2 = 1 	imes 10^{-3} 	imes 3.36 	imes 10^5 = 336 \ J$.Total heat $Q = Q_1 + Q_2 = 420 \ J$.$10.5m + 336 = 420$.$10.5m = 420 - 336 = 84$.$m = 84 / 10.5 = 8$.Therefore,the value of $m$ is $8 \ g$.

Similar Questions

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