Question

How much heat energy is required to change a 40 g ice cube from a solid at -10 oC to steam at 110 oC?​

Answer 1

Answer:

I would like to use calories in my solution and then convert to joules. I will take the specific heats of ice and steam as about 0.5 cal/g-K. Latent heat of melting is 80 cal/g and for vaporization is 540 cal/g

Heat needed to raise ice to melting: 40 * 0.5 * 10 = 200 cal

heat needed to melt ice: 40 * 80 = 3200 cal

heat to raise water to 100: 40 * 1 * 100 = 4,000 cal

heat to evaporate water: 40 * 540 = 21600 cal

heat to raise to 110: 40 * 0.5 * 10 = 200 cal

total: 29,200 cal = 1.22 E 5 Joules