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.
I think Bob Pursley worked this problem for you but you don't give all of the data needed here. Here is how to do it.
q1 = heat to move CH3OH from 20 to 30.
q1 = mass CH3OH x spcific heat CH3OH x (delta T).
q2 =heat to vaporize CH3OH at 30.
q2 = mass CH3OH x heat of vaporization at 30.
Then total heat = q1 + q2.
I must point out that CH3OH boils at about 65 or so so the usual heat of vaporization at 65 may not be the same as the heat of vaporization at 30. If the information in your original post was not the same then he may have worked the problem somewhat differently.
hm...I don't remember posting this problem but thank you again dr.bob, I can definetly always count on you
Btw what do I do about the molar heat capacity
To calculate the amount of heat required to raise the temperature of CH3OH, followed by vaporization, you need to consider two separate steps:
Step 1: Calculate the amount of heat required to raise the temperature of CH3OH from 20 to 30 degrees Celsius.
To calculate the heat required for this step, you need to use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity (in J/g·K)
ΔT is the change in temperature (in Celsius)
Given:
m = 200 g (mass of CH3OH)
c = 81.1 J/g·K (specific heat capacity of CH3OH)
ΔT = 30°C - 20°C = 10°C
Now, substitute the values into the formula:
Q1 = 200 g * (81.1 J/g·K) * 10°C
Calculating this, you get:
Q1 = 162,200 J
Step 2: Calculate the amount of heat required for the vaporization of CH3OH at 30 degrees Celsius.
To calculate the heat required for this step, you need to use the formula:
Q = m * ΔHvap
Where:
Q is the heat energy (in Joules)
m is the mass of the substance (in grams)
ΔHvap is the enthalpy of vaporization (in J/g)
For CH3OH, the enthalpy of vaporization is not given in your question. In order to proceed, you need to find this value from a reliable source or your textbook. Let's assume ΔHvap is 40,000 J/g.
Given:
m = 200 g (mass of CH3OH)
ΔHvap = 40,000 J/g (assumed enthalpy of vaporization)
Now, substitute the values into the formula:
Q2 = 200 g * 40,000 J/g
Calculating this, you get:
Q2 = 8,000,000 J
Finally, calculate the total amount of heat required by adding Q1 and Q2 together:
Total heat required = Q1 + Q2
Total heat required = 162,200 J + 8,000,000 J
Total heat required ≈ 8,162,200 J
So, the approximate amount of heat required to raise the temperature of 200 g CH3OH from 20 to 30 degrees and then vaporize it at 30 degrees is around 8,162,200 Joules.