Question

You have 750 g of ice at -20 degrees celsius. We are going to calculate how much energy it takes to evaporate the ice and raise it to a temperature of 120 degrees celsius. Mark the initial and final temperat

Answer 1

heat absorbed = m x c x ∆T

∆T = 120-(-20) = 140°c

m = 750 g

c=1cal/g°c

heat needed = 750 x 140 x 1

= 10500 cal

however latent heat is also required

=540 cal/g ....latent heat of vaporisation of water

latent heat req for 750g

= 540 x 750

= 40500 cal

latent heat of fusion

= 80 cal/ g

for 750g ..... 80*750 = 60,000

Total heat req = 60,000+10500+40500 cal

= 111000 cal

or 111 kcal

Answer 2

20*225*750 + 120*335*750,