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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?

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2 Answers

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.

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