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have this problem here from chapter 5 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 one kilogram one 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 right raising the liquid temperature raising raising liquid just raising the liquid temperature let's figure out how we're going to do that just the liquid temperature so the first thing I like this is well how many degrees are we raising the temperature well we're going from 20 degrees Celsius to so we're going from 20 to 20 degrees Celsius to 70 let me write Celsius they're just so weak it's clear 20 degrees Celsius to 78.29 degrees Celsius so how much did we raise it well 78.29 minus 20 is fifty eight point two nine so our change in temperature our change in temperature is so our change in temperature is equal to fifty eight point two nine degrees and we could actually fifty eight point two nine degrees Celsius or this could even be fifty eight fifty eight point two nine Kelvin and the reason why we can do that is because difference is on the Celsius scale and the Kelvin kiss scale are the same thing the scale the Cal the Kelvin scale is just a scaled is just a shifted version of the Celsius scale if you added - 73 - 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 - 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 5850 8.29 of staines Celsius 0.29 actually let me change it to Kelvin because that looks like what our units are given in terms of specific heat so let me write that down 50 8.29 Kelvin is our change in temperature I could have converted either of these to kelvin first then found the difference gotten the exact same number because the Celsius scale in the Kelvin scale the increments are the same amount now that's our change in temperature now how much how much ethanol are we trying to boil well it tells us right over here tells us that we're dealing with one one kilogram one kilogram of ethanol and everything else they gave us is in gram so let me just write that 1 kilogram that's the same thing as a thousand grams we could just write it here 1.000 kilogram is equal to that or let me write it this way times 1,000 grams per 1 kilogram these cancel out this is the same thing as one thousand one thousand grams although the reality is we only have three significant digits this makes it look like we have four so we have one thousand grams times one thousand grams and then we just multiply this times the specific heat of ethanol the specific heat capacity of ethanol right here two point four four joules per gram Kelvin so times two point four four joules let me write it this way two point four four joules per gram per gram Kelvin you see that the units work out this Kelvin is going to cancel out with that company in the denominator 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 you that is required to push it at one degree so here we're doing 58 degrees a thousand grams you just multiply the units cancel out so you have Kelvin canceling out with Kelvin you have grams cancelling out with grams and we are left with take out the calculator we take out put it on the side here so we have fifty eight point two nine times one thousand times one two three one thousand times two point four four is equal to and we only have three significant digits here so this is only this is going to be one hundred forty two we'll just round down 142,000 Kelvin so this is one hundred and forty mm sorry 140 mm Jules Jules is our units we want energy so this is how much so this right here is the amount of energy to take our ethanol our one 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 point two nine Kelvin now the next step is it's just a lot warmer ethanol liquid ethanol we now have to make it we now have to vaporize it has to become vapor at that temperature so now we just add we have to add the heat of vaporization so that's right here or we 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 or a 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 a thousand grams we know we have a thousand grams a thousand grams of ethanol the grams cancel out 855 times 1,000 is 800 and thirty-five thousand joules so it actually took it actually took a lot less energy to make the ethanol go from twenty degrees Celsius to seventy eight point two nine degrees Celsius then it took it to stay at seventy eight point two nine 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 we could actually let's see if we can add this up in our heads eight hundred fifty five thousand plus eight hundred forty two thousand eight hundred plus 100 is nine hundred its nine hundred thousand fifty plus forty is ninety five plus two is seven so it's nine hundred and ninety not nine hundred ninety seven thousand joules or nine hundred ninety seven kilojoules or we could almost say almost it's almost one mega Joule if we wanted to speak in those terms but that's what will take for us to vaporize that one kilogram of ethanol