At 1 atm, how much energy is required to heat 81.0 g H2O(s) at −12.0 ∘C to H2O(g) at 125.0 ∘C? Use the heat transfer constants found in this table.

Question

	Quantity	per gram
Enthalpy of fusion	333.6 J/g	6010. J/mol
Enthalpy of vaporization	2257 J/g	40660 J/mol
Specific heat of solid H2O (ice)	2.087 J/(g·°C) *	37.60 J/(mol·°C) *
Specific heat of liquid H2O (water)	4.184 J/(g·°C) *	75.37 J/(mol·°C) *
Specific heat of gaseous H2O (steam)	2.000 J/(g·°C) *	36.03 J/(mol·°C) *

I got 2.77 e2 for my answer, but apparently I need three sig figs, and 277K is also wrong?

Answer 1

There are several steps that should be taken to answer this question.

(1). Heat needed to raise temp. of 81 g solid H2O from -12º to 0º - use sp. heat of ice

(2). Heat needed to melt 81 g solid H2O to liquid H2O at 0º - use ∆Hfusion

(3). Heat needed to raise temp. of 81 g liquid H2O from 0º to 100º - use sp. heat of liquid H2O

(4). Heat needed to convert 81 g liquid water to gaseous water at 100º - use ∆Hvap

(5). Heat needed to raise temp of 81 g of steam from 100º to 125º - use sp. heat of steam

(1). q = mC∆T = (81 g)(2.087 J/g/deg)(12 deg) = 2029 J

(2). q = m∆Hfus = (81 g)(333.6 J/g) = 27,022 J

(3). q = mC∆T = (81 g)(4.184 J/g/deg)(100 deg) = 33,890 J

(4). q = m∆Hvap = (81 g)(2257 J/g) = 182,817 J

(5). q = mC∆T = (81 g)(2.00 J/g/deg)(25 deg) = 4050 J

Add up all the joules to get 249,808 J = 249.8 kJ = 250. kJ (to 3 significant figures; note the decimal)