Question

How much heat energy or heat is required to change a 45 g ice cube at -120C into steam at 1120C.

Answer 1

The required amount of heat

$$Q=Q_1+Q_2+Q_3+Q_4+Q_5$$

$$Q=c_{ice}m\Delta T_1+\lambda m\\ +c_{water}m\Delta T_2+Lm+c_{steam}m\Delta T_3$$

$$Q=2093*0.045*12+333550*0.042\\ +4186*0.042*100+2260000*0.042\\ +2010*0.042*12=1.29*10^5\:\rm J$$