A 100.0 g sample of ethanol at 25.0 degrees celsius is heated and boiled completely to vapor at its boiling point. How much heat does it absorb?
i did....
q=100.0 *2.44*43.5=10,614 J
q=100.0*2.44*53.5=13,054.
then i added the two q's together and i got 23,700 J. is that right??? if no can someone please tell me what to do?
You need to tell me what numbers you are using. The second q of 13054 looks ok but I used 78.4 for the boiling point and 78.4-25 = 53.4). Perhaps you used a different boiling point.
I found several values for heat of vaporization. You should have mass x heat vaporization, then add the two qs together. I don't know what the 43.5 is in the first q but probably that part is wrong. Perhaps I can be more helpful if you tell me what values you are using.
43.5 i got off a phyical state changing chart...
If its 45.3 J/g*C for the heat of vaporization, then q = 100 x 45.3 = ??
However, I didn't see anything that looked like that when I looked for delta H vap.
To calculate the heat absorbed by a substance during a phase change, you need to use the formula:
q = m * ΔH
Where:
q is the heat absorbed or released (in Joules)
m is the mass of the substance (in grams)
ΔH is the enthalpy of the substance (in J/g)
For ethanol, the enthalpy of vaporization (ΔH) is 38.56 J/g.
Given:
m = 100.0 g (mass of ethanol)
ΔH = 38.56 J/g (enthalpy of vaporization)
Now, let's calculate the heat absorbed by the ethanol using the above formula:
q = 100.0 g * 38.56 J/g
q = 3,856 J
Therefore, the correct answer is that the ethanol absorbs 3,856 J of heat.