At 1 atm, how much energy is required to heat 81.0 g H2O(s) at −20.0 ∘C to H2O(g) at 119.0 ∘C?

Answer 1

It's best to do these problems in steps. We also need certain constants. The values I use may vary slightly from the ones you find. I am using the following:

C_ice = 2.09 J/gº

C_{liq water} = 4.184 J/gº

C_steam = 1.89 J/gº

∆H_fusion = 334 J/g

∆H_vap = 2260 J/g

(1) heat to raise temp of 81.0 g of ice from -20º to 0º

q = mC∆T = (81 g)(2.09 J/gº)(20º) = 3386 J

(2) heat to melt 81 g of ice @ 0º (phase change)

q = m∆H_fusion = (81 g)(334 J/g) = 27,054 J

(3) heat to raise temp of 81 g of liquid water from 0º to 100º

q = mC∆T = (81 g)(4.184 J/gº)(100º) = 33,890 J

(4) heat to change 81 g liquid water to steam @ 100º (phase change)

q = m∆H_vap = (81 g)(2260 J/g) = 183,060 J

(5) heat to raise the temp of steam from 100º to 119º

q = mC∆T = (81 g)(1.89 J/gº)(19º) = 2910 J

Add up all the joules to get the total energy needed...

3386 + 27054 + 33890 + 183060 + 2910 = 250,300 J = 250.3 kJ