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?
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.
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
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).
The answer on mast chem is 279 if you have 230 grams instead of 200.
To calculate the amount of heat required to raise the temperature of a substance and then vaporize it, you need to consider two separate steps:
Step 1: Heating the substance and raising its temperature from 20°C to 30°C.
Step 2: Vaporizing the substance at 30°C.
Let's break down the steps and calculate the heat required:
Step 1: Heating the substance using the formula: q = m × C × ΔT
1. Convert the given mass from grams to moles:
m(CH3OH) = 200 g
Molar mass of CH3OH = 32.04 g/mol
moles(CH3OH) = m(CH3OH) / Molar mass
moles(CH3OH) = 200 g / 32.04 g/mol
moles(CH3OH) ≈ 6.237 mol
2. Calculate the temperature change:
ΔT = T2 - T1
ΔT = 30°C - 20°C = 10 K
3. Plug the values into the formula to calculate the heat required:
q1 = moles(CH3OH) × C × ΔT
q1 = 6.237 mol × (81.1 J/mol/K) × 10 K
q1 ≈ 5067.53 J
Step 2: Vaporizing the substance using the formula: q = m × ΔHvap
1. The heat required to vaporize 1 mole of CH3OH is called the molar enthalpy of vaporization (ΔHvap). In this case, we need to find the value of ΔHvap for CH3OH.
2. Assuming you don't have the value of ΔHvap, you are unable to directly calculate the heat required for vaporization without this information. Therefore, you need to consult your textbook or any other reliable source to find the value of ΔHvap for CH3OH. Once you have that information, you can calculate q2 using the formula mentioned above.
Please note that the value of 254 you mentioned might include the heat required for vaporization (q2) but without knowing the value of ΔHvap, we can't confirm if that is correct.