States of matter
Change of state example Specific Heat Capacity and Enthalpy of Vaporization example
Change of state example
- I have this problem here from chapter five of the Kotz,
- Treichel, and Townsend Chemistry and Chemical
- Reactivity book, and I'm doing this with their permission.
- So they tell us that ethanol, C2H5OH, boils at-- let me do
- this in orange-- it boils at 78.29 degrees Celsius.
- How much energy, in joules, is required to raise the
- temperature of 1 kilogram of ethanol from 20 degrees
- Celsius to the boiling point and then change the liquid to
- vapor at that temperature?
- So there's really two parts of this problem.
- How much energy, in joules, to take the ethanol from 20
- degrees to 78.29 degrees Celsius?
- That's the first part.
- And then once we're there, we're going to have 78.29
- degrees Celsius liquid ethanol.
- But then we also need the energy to turn it into vapor.
- So those are going to be the two parts.
- So let's just think about just raising the liquid
- Raising, Raising the liquid temperature
- Let's figure out how we're going to do that.
- Just the liquid temperature.
- So the first thing I looked at is well how many degrees are
- we raising the temperature?
- Well we're going from 20 degrees Celsius-- let me write
- Celsius there just so it's clear-- 20 degrees Celsius to
- 78.29 degrees Celsius.
- So how much did we raise it?
- Well, 78.29 minus 20 is 58.29.
- So our change in temperature is equal to 58.29 degrees
- Celsius or this could even be 58.29 kelvin.
- And the reason why we can do that is because differences on
- the Celsius scale and the kelvin scale
- are the same thing.
- The kelvin scale is just a shifted version
- of the Celsius scale.
- If you added 273 to each of these numbers you would have
- the kelvin temperature, but then if you take the
- difference, it's going to be the exact same difference.
- Either way you do it, 78.29 minus 20.
- So that's how much we have to raise the temperature.
- So let's figure out how much energy is required to raise
- that temperature.
- So we want a delta T.
- We want to raise the temperature 58.29.
- I'll stay in Celsius.
- Actually let me just change it to kelvin because that looks
- like what our units are given in terms of specific heat.
- So let me write that down.
- 58.29 kelvin is our change in temperature.
- I could have converted either of these to kelvin first, then
- found the difference, and gotten the exact same number.
- Because the Celsius scale and the kelvin scale, the
- increments are the same amount.
- Now, that's our change in temperature.
- Now how much ethanol are we trying to boil?
- Well, it tells us right over here.
- It tells us that we're dealing with 1 kilogram of ethanol.
- And everything else they give us is in grams. So let me just
- write that 1 kilogram, that's the same thing as 1,000 grams.
- We could just write it here.
- 1.00 kilogram is equal to-- or let me write it this way--
- times 1,000 grams per 1 kilogram.
- These cancel out.
- This is the same thing as 1,000 grams. Although the
- reality is we only have three significant digits-- this
- makes it look like we have four.
- So we have 1,000 grams times 1,000 grams. And then we just
- multiply this times the specific heat of ethanol.
- The specific heat capacity of ethanol right here, 2.44
- joules per gram kelvin.
- So times 2.44 joules.
- Let me write it this way, 2.44 joules per gram kelvin.
- You see that the units work out.
- This kelvin is going to cancel out with that kelvin in the
- This gram in the numerator will cancel out with that
- grams. And it makes sense.
- Specific heat is the amount of energy per mass per degree
- that is required to push it that 1 degree.
- So here we're doing 58 degrees, 1,000 grams, you just
- multiply it.
- The units cancel out.
- So you have kelvin canceling out with kelvin.
- You have grams canceling out with grams.
- And we are left with-- take out the calculator, put it on
- the side here.
- So we have 58.29 times 1,000-- times one, two, three-- times
- 2.44 is equal to-- and we only have three
- significant digits here.
- So this is going to be 142-- we'll just round down--
- 142,000 kelvin.
- So this is 142,000.
- Sorry 142,000 joules.
- Joules is our units.
- We want energy.
- So this right here is the amount of energy to take our
- ethanol, our 1 kilogram of ethanol, from 20 degrees
- Celsius to 78.29 degrees Celsius.
- Or you could view this as from 293 kelvin to whatever this
- number is plus 273, that temperature in kelvin.
- Either way, we've raised its temperature by 58.29 kelvin.
- Now, the next step is, it's just a lot warmer ethanol,
- liquid ethanol.
- We now have to vaporize it.
- It has to become vapor at that temperature.
- So now we have to add the heat of vaporization.
- So that's right here.
- We should call it the enthalpy of vaporization.
- The enthalpy of vaporization, they tell us, is
- 855 joules per gram.
- And this is how much energy you have to do to vaporize a
- certain amount per gram of ethanol.
- Assuming that it's already at the temperature of
- vaporization, assuming that it's already at its boiling
- point, how much extra energy per gram do you have to add to
- actually make it vaporize?
- So we have this much.
- And we know we have 1,000 grams of enthanol.
- The grams cancel out.
- 855 times 1,000 is 855,000 joules.
- So it actually took a lot less energy to make the ethanol go
- from 20 degrees Celsius to 78.29 degrees Celsius than it
- took it to stay at 78.29, but go from the liquid form to the
- vapor form.
- This took the bulk of the energy.
- But if we want to know the total amount of energy, let's
- see if we can add this up in our heads.
- 855,000 plus 142,000.
- 800 plus 100 is 900.
- That's 900,000.
- 50 plus 40 is 90.
- 5 plus 2 is 7.
- So it's 997,000 joules or 997 kilojoules.
- Or we could say it's almost 1 megajoule, if we wanted to
- speak in those terms. But that's what it will take for
- us to vaporize that 1 kilogram of ethanol.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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