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At 1 atm, how much energy is required to heat 77.0 g of H2O(s) at –12.0 °C to H2O(g) at 121.0 °C?

I've tried this process over and over again and keep getting extremely high numbers... what am I doing wrong?

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) * 


What did you get for an answer, Kayla?

2 Answers by Expert Tutors

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Randall K. | Experienced Chemistry TutorExperienced Chemistry Tutor
4.9 4.9 (187 lesson ratings) (187)

Hello Kaya,


Using your numbers for the specific heats of solid, liquid, and gaseous water and the enthalpies of fusion and vaporization, I get 236.8 kJ of energy.


Starting at -12.0 degrees C,

q1=m CP dt=77.0g(2.087 J/g*C)(0+12.0 C)=1.93 kJ

q2=dH of fusion=6010 J/mol(77.0 g)(1 mol/18.02 g)=25.7 kJ

q3=m CP dt=77.0 g(4.184 J/g*C)(100. C-0 C)=32.2 kJ

q4=dH of vaporization=40660 J/mol(77.0g)(1 mol/18.02 g)=173.7 kJ

q5= m Cp dt=77.0g(2.000 J/g*C)(121.0 C-100. C)=3.23 kJ

The sum of q1-q5=236.8 kJ


Hope this helps.



Given that the measured value for water is 77.0 g (three significant figures), the answer should be reported as 237 kJ.

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)

From -12 Co to 0 Co (ice), Q1 = mCp1*?T = 77*2.087*12 J

Heat of melting: Q2 = mL2 = 77*333.6 J

From 0 Co to 100 Co (water), Q3 = mCp3*?T = 77*4.184*100 J

Heat of evaporation: Q4 = mL3 = 77*2257 J

From 100 Co to 121 Co (steam), Q5 = mCp5*?T = 77*2.000*21 J

Q = Q1+Q2+Q3+Q4+Q5 = 2.369 x 105 J <==Answer