At 1 atm, how much energy is required to heat 51.0 g of H2O(s) at –12.0 °C to H2O(g) at 119.0 °C?

Answer 1

To answer this question, you should look at the various steps involved in going from water as a solid (ice) at -12ºC, to water as a gas (steam) at 119.0ºC. These steps are:

(1) raise the temperature from -12ºC to 0ºC

(2) melt the solid water at 0ºC (phase change)

(3) raise temperature of liquid from 0ºC to 100ºC

(4) convert the liquid to a gas at 100ºC (phase change)

(5) raise temperature of gas from 100ºC to 119ºC

(6) Add up all the heat needed in the different steps

For the following calculations, I'm using

C_ice = 2.09 J/g/deg

C_l = 4.184 J/g/deg

C_g = 1.89 J/g/deg

∆H_f = 334 J/g

∆H_vap = 2260 J/g

(1) q = mC∆T = (51.0g)(2.09 J/g/deg)(12 deg) = 1279 J

(2) q = m∆H_f = (51.0g)(334 J/g) = 17,034 J

(3) q = mC∆T = (51.0g)(4.184 J/g/deg)(100 deg) = 21,338 J

(4) q = m∆H_vap = (51.0g)(2260 J/g) = 115,260 J

(5) q = mC∆T = (51.0g)(1.89 J/g/deg)(19 deg) = 1831 J

∑steps 1-5 = 156,742 J = 157 kJ