I can answer part of the question but not all of it.
First, convert 200 g CH3OH to moles.
Then heat required to heat that many moles of CH3OH will be
moles x heat capacity x delta T =
(200/32) x 81.1 J/mol x (30-20) = a much larger number than you are giving for the answer AND there is more to add.
The above comes from
q = mass x specific heat x (Tfinal-Tinitial)
You can use mass of 200 and convert the 81.1 J/mol to J/gram or you can convert 200 g to moles and use the 81.1 as is. In any event, we have an answer close to 5000 J just for heating the 10 degrees C. The questions asks for the stuff at 30 C to be vaporized yet you provide no heat of vaporization at that temperature. Whatever that value is must be added to the 5000 J already calculated to arrive at the total q.
With a value of 254, no units, the answer could be anything.
How much heat is required to raise the temperature of 200 g CH3OH from 20 to 30 degrees and then vaporize it at 30 degrees. The molar heat capacity of CH3OH is 81.1 J/Mol/K.
Okay...the answer is approximately around 254. It was a homework problem. The question is how do you get to the answer? I had someone give me the answer, but I want to know how to do it. So can anyone just show me the steps to get to the answer.
Redirected to dr. bob or anyone else, what do I do with the molar heat capacity?
5 answers
Heat of vap. is 38.9 KJ/mol
Then moles in 200 = 200/32 = ??
total heat =
[moles x 81.1 x (10)] + [moles x 38,900] = ??
I get about 248,194 J or about 248 kJ.
That should be the correct answer if 38.9 kJ is the heat of vaporization AT 30 C (and not at it's boiling point). (38.9 sounds high to me but go with what ever is
total heat =
[moles x 81.1 x (10)] + [moles x 38,900] = ??
I get about 248,194 J or about 248 kJ.
That should be the correct answer if 38.9 kJ is the heat of vaporization AT 30 C (and not at it's boiling point). (38.9 sounds high to me but go with what ever is
Then moles in 200 = 200/32 = ??
total heat =
[moles x 81.1 x (10)] + [moles x 38,900] = ??
I get about 248,194 J or about 248 kJ.
That should be the correct answer if 38.9 kJ is the heat of vaporization AT 30 C (and not at it's boiling point). (38.9 sounds high to me. Have you used the Clausius-Clapeyron equation? If so you can use the CC equation to calculate the heat of vaporization at 30 C, then substitute it in the above work).
total heat =
[moles x 81.1 x (10)] + [moles x 38,900] = ??
I get about 248,194 J or about 248 kJ.
That should be the correct answer if 38.9 kJ is the heat of vaporization AT 30 C (and not at it's boiling point). (38.9 sounds high to me. Have you used the Clausius-Clapeyron equation? If so you can use the CC equation to calculate the heat of vaporization at 30 C, then substitute it in the above work).
The answer on mast chem is 279 if you have 230 grams instead of 200.
3 answers