Answer to Question #204330 in Physics for Morgan Hopkin

Question

Find the energy required for Kamryn to vaporize 0.65 kg of gold. (mLv)

Answer 1

Assume that the initial temperature of gold is $t_0=0°$ .

$Q_1=cm\Delta t=129\cdot0.65\cdot1064=89265\ (J)$

$Q_2=\lambda m=66200\cdot0.65=43030\ (J)$

$Q_3=rm=1736340\cdot0.65=1128621\ (J)$

$Q=Q_1+Q_2+Q_3=89265+43030+1128621=1260916\ (J)$ . Answer