posted by Anonymous .
At 1 atm, how much energy is required to heat 67.0 g of H2O(s) at –14.0 °C to H2O(g) at 115.0 °C?
Please explain, I'm not very good at Chemistry and this problem is giving me a headache. Thank you!
There are two equations you need and you do each in steps.
For moving from one T to another in the same phase, the formula is
q = heat needed = mass H2O x specific heat H2O x (Tfinal-Tinitial). For example, to move 67.0 g water from zero C to 100 C, heat needed is
q = 67.0 x 4.184 J/g*C x (100-0) = ?. Notice what I mean by the same phase. It is liquid from zero C to 100 C.
The other equation you need is ast the phase change. There are two phase changes in water for the temperatures in the problem. From ice at zero C to liquid H2O at zero C it is
q = mass H2 x heat fusion.
The other phase change is at 100 C. It changes from liquid water to steam at 100 C. That one is the same form but is
q = mass H2O x heat vaporization.
So you have same phase from -14 C to zero C(that's all ice phase or solid). Then a phase change at zero C from solid to liquid. Same phase from zero C to 100 C (all liquid). Phase change at 100 C to steam. Then same phase from 100 C steam to 115 C steam. Then add all of the qs together.
Any of the heat problems like this are worked the same way. Those two equations will work hundreds of questions.