Question

How much energy is required to change a 40-g ice cube from ice at -10 Celsius to steam at 110 Celsius?

Answer 1

Solution- As we know ice will turn to water at 0 C, ΔT1 = 10
So,
Q1 = mciceΔT1 + mLfusion

= 40 x 2.108 x 10 + 40 x 334.0

Q1 = 14203.2 J

Now again steam at 100 C, ΔT2 = 100

Q2 = mcwaterΔT2 + mLvaporisation

= 40 x 4.187 x 100 + 40 x 2260

Q2 = 107148 J

And at last, to the end point ΔT3 = 10

(here's no phase change)

Q3 = mcsteamΔT3

= 40 x 1.996 x 10

Q3 = 798.4 J

Now we just add all the three to get total
QTotal = Q1 + Q2 + Q3

= 14203.2 + 107148 + 798.4

= 122149.6 J

= 122.15 kJ