The specific heat value for water doesn't change. It still is 4.18 J/g*C.
And yes, you still must use q = mc(delta T)
What you have to do in these cases, where obviously there is some water left over, is to go about it somewhat backwards.
q = mass x specific heat x (Tfinal-Tinitial)
Calculate how much heat is used in raising the T of 35.00 grams H2O from 30 C to 100 C. That will be
q = 36.00 x 4.18 x (100-30) = approx 10,000 J. That will leave you how much heat not used? That is 30,000-10000 = approx 20000 J remaining which can be used to turn some of the water at 100 C into steam.
Calculate how many g that will handle.
q = mass x heat vaporization
20,000 = mass x 2260 J/g
Solve for mass H2O = approx 9 grams.
We had 35 g H2O to start; approx 9 have been transformed into steam, you have left approx 35-9 = ?
Note: I hav estimated all of the numbers; you will need to recalculate each step more accurately than I've done.
A 35.00 mL sample of water at 30.0°C is placed in an open container with an external pressure of 1.00 atm.
30.0 kJ of heat is added to the water sample. The density of water is 1 g/mL.
(a) How many grams of liquid water remain at the end of the process?
Like, the equation I was going to use was
q = C*(delta)T with C being the specific heat for water (or 4.18 J/(g*degrees Celsius))
However, then I have the problem of adding heat to the equation, and not really knowing where to go from there (can you still use the 4.18 J/(g*degrees Celsius) as your specific heat value? How are you supposed to find temperature change if you don't know what specific heat value to use? How on earth do you figure out how much water is left at the end of all this?)
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