lilsis76
  • lilsis76
I need help understand, and doing this problem: Calculate the amount of energy it will take to vaporize 45.7 grams of SOLID water at 255K to 380K.
Chemistry
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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lilsis76
  • lilsis76
okay let me try
lilsis76
  • lilsis76
AHH!!!! @aaronq I dont know how you got 255C if its 255 K. Okay so I did 255K-273=-18 deg Celsius and 380-273= 107 deg Celsius
aaronq
  • aaronq
damn, i mean K :S sorry. do you have the value for the enthalpy of sublimation?

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lilsis76
  • lilsis76
ugh.... no. this is a practice take home test cuz our exam is tuesday and this is all I have
aaronq
  • aaronq
we'll you would multiply the mass (or moles, depending what units \(\Delta H_{sublimation}^{255K}\) is in), then find the energy req'd to change the temp from 255 K to 380 K.
lilsis76
  • lilsis76
what? sorry im a visual learner umm... okay well i have 45.7 g of Sol. Water at 255K-273= -18 celsius 380K-273= 107 celsius
aaronq
  • aaronq
okay sorry, i interpreted it wrong, so you need to do this in 4 steps. |dw:1378516650982:dw| for steps 1, 3 and 5 you're using the formula: \(q=m*C*\Delta T\) what differs is the specific heat capacity (C), because it's different when it's ice or water or vapour. for steps 2 and 4 you're using the formula: \(q=\Delta H_{y}*m\) (where y=fusion for step 2 and y=vaporization for step 4). After you found each individual one, add all the q's together and that's your final answer.
aaronq
  • aaronq
i meant 5 steps not 4****
lilsis76
  • lilsis76
okay let me erase what i have. I had a bunch of dumb numbers And i dont know how i ended up with a -32.7
lilsis76
  • lilsis76
haha okay @aaronq let me see well for the 1) 45.7g ( I have to get it to ....kJ or Joules Right?
aaronq
  • aaronq
yeah you have to find the heat (which can be in any units of energy) to raise the temp of the ice to 0 degrees celsius (or 273 K)
lilsis76
  • lilsis76
so.....45.7g( 1mol/18gh2o) so it would be 2.53 mols? am i in the right direction? @aaronq
aaronq
  • aaronq
you only need to convert to moles if the units of the specific heat capacity are in "J/mol*K" but they're normally in grams"J/g*K", so that conversion is not necessary.
lilsis76
  • lilsis76
oh.
lilsis76
  • lilsis76
well then I have nothing else to go by :(
aaronq
  • aaronq
okay so first you should find the values of the specific heat capacities for ice, water and water vapour. Also, find the Enthalpy of fusion (of ice) and Enthalpy of vaporization. Write these down somewhere because they're useful.
lilsis76
  • lilsis76
okay im writing now
lilsis76
  • lilsis76
okay im going to search ill be back in a couple mintues
aaronq
  • aaronq
okay, you're gonna use those values for all these questions, so try to find the in the same units (J/g*K) .. kelvin or celsius, it doesn't matter
lilsis76
  • lilsis76
okay so far i found these: ice warming-->.941 kJ/mol ice melt -->6.02 kJ/mol Liquid water warm-->7.52 kJ/mol Liquid vaporizing-->40.7 kJ/mol steam warming-->.904 kJ/mol
lilsis76
  • lilsis76
I dont have enthalphy fusion, or enthalpy vaporization in my book :( @aaronq
aaronq
  • aaronq
enthalpy of fusion = ice melt -->6.02 kJ/mol enthalpy of vaporization = Liquid vaporizing-->40.7 kJ/mol
aaronq
  • aaronq
okay, since all your values are in moles, you do need to convert grams to moles.
lilsis76
  • lilsis76
OH 0.o I guess i put it but I didnt recognize it as entalpys
lilsis76
  • lilsis76
okay, im ready. whats next
aaronq
  • aaronq
okay so step 1) find the heat req'd to raise the temp of the ice to 273 K (zero degrees celsius) use: \(q=m*C*(T_f-T_i)\)
lilsis76
  • lilsis76
okay so for number 1: q= m C A T would be....... 45.7g( 6.02 kJ/mol) ( Tfinal -T initial )
lilsis76
  • lilsis76
right? it looks right but I think its missing alot
aaronq
  • aaronq
you're using: ice warming-->.941 kJ/mol = C because the ice is warming up not melting whats your initial temp and whats your final temp for this step?
lilsis76
  • lilsis76
i just have the 255K and 380K
aaronq
  • aaronq
but you're only warming up the ice to it's melting point.
aaronq
  • aaronq
look at the graph i drew
lilsis76
  • lilsis76
okay so..... i have.. okay so... We have a solid ice at... 255K a liquid at 273K
lilsis76
  • lilsis76
okay @aaronq so were only staying between the temperatures of 255K to373K since the ice is a sold, then its liquid. right?
aaronq
  • aaronq
yes, but it's still solid, what were calculating in this step is the heat req'd to raise the temp of the ice (not melting just yet, thats step 2). but yes, so whats \(T_f\) and what's \(T_i?\)
lilsis76
  • lilsis76
Im so lost @aaronq :( I cant figure out how to start it
aaronq
  • aaronq
okay you're starting at 255 K right? so thats \(T_i\), the initial temperature. and you're going to 373 K (the melting temp of ice), so this is \(T_f\), can you plug it into the equation now?
lilsis76
  • lilsis76
i think i have to use the .941kJ/ 1mol and the 45.7 grams right?
aaronq
  • aaronq
yes and no, you have to use moles because the specific heat capacity (0.941 kJ/mol) is in moles.
lilsis76
  • lilsis76
ok ok so is it like. this? hold on
lilsis76
  • lilsis76
45.7g* \[(\frac{ .941 J }{ 1 mol }) (373k-255k)\]
aaronq
  • aaronq
remember the units of C are in moles, so you need to convert 45.7 g to moles but you're on the right track
lilsis76
  • lilsis76
how do I do that again? the grams to moles? is that.... 45.7g(1mol/18.02gwater)=2.54moles water?
aaronq
  • aaronq
yep thats right.
lilsis76
  • lilsis76
YAY! i did 1 thing right! :D
aaronq
  • aaronq
haha woo ! now step 2 !
lilsis76
  • lilsis76
okay okay. so now..... i.... haha okay hold on I gotta um 2.54moles(.941 kJ/mol)(373k-255k)(45.7grams Solid water) right??? :D ???
aaronq
  • aaronq
you don't need to use the mass, because you're already using its equivalent value in moles. "2.54moles(.941 kJ/mol)(373k-255k)=q" is good
aaronq
  • aaronq
so step 2?
lilsis76
  • lilsis76
sorry said no connection haha
lilsis76
  • lilsis76
okay @aaronq Im calculating and got this.. 282.03 is it K or kJ?
aaronq
  • aaronq
K is Kelvin a unit of temperature, kJ is kilojoules (1000 joules). We're finding heat energy here
lilsis76
  • lilsis76
okay were finding heat energy. so it would be in kJ so the answer is 282.03 kJ right?
aaronq
  • aaronq
\(\checkmark\)
lilsis76
  • lilsis76
YAY!!! I did good for part 1 right? :D
aaronq
  • aaronq
yes! so write down \(q_1=282.03 kJ\) because you're gonna need it at the end.
aaronq
  • aaronq
step 2: \(q_2=\Delta H_{fusion}*n\)
lilsis76
  • lilsis76
yes I did okay PART 2! its q= \[\Delta H(m)\]
lilsis76
  • lilsis76
is n the number we just got?
lilsis76
  • lilsis76
okay @aaronq q=delta H fus (n) q= 6.02 kJ/mol (282.03 kJ ) right?
aaronq
  • aaronq
n= number of moles the \(q_1\) is heat and you don't need to use it until you add all of the q's up.
aaronq
  • aaronq
n= moles of H2O
lilsis76
  • lilsis76
OH! okay so thats the 18.02g
lilsis76
  • lilsis76
so its....6.02(18.02)=108.48 ??? @aaronq and Ill be right back mom wants me to drop her off at my aunts really fast. will only take me 5 minutes ill be right back. but in the mean time did i do it right?
aaronq
  • aaronq
that's okay, i'll probably be here. and thats correct! but remember to always write units, because otherwise those values have no meaning.
lilsis76
  • lilsis76
okay im back. sorry longer than expected. I gotta pick mom up at 815 so in about .....lol almost 15 min, yay me :/ I hate driving haha Okay ill write... 6.02 kJ/mol (18.0 g h20) = 108.48 g*kJ/mol okay were on to part 3 right?
aaronq
  • aaronq
no worries ! lets do this quick then. yep, part 3!
lilsis76
  • lilsis76
okay., part 3 Is 1=mC Delta T
aaronq
  • aaronq
now that you've melted the ice, you gotta raise the temp of the water to it's boiling point (373 K)
aaronq
  • aaronq
yep, but since C is in kJ/mol, you need to substitute m for n so q=n*C*dT
aaronq
  • aaronq
essentially, what I'm saying is use moles not mass
lilsis76
  • lilsis76
okay
lilsis76
  • lilsis76
2.54moles(7.52 kJ/mol)( Tfinal-373) what do i put for Tfinal? @aaronq
aaronq
  • aaronq
the temperatures are incorrect, you melted the ice, right? so that was 273 K boiling point of water 373 K
lilsis76
  • lilsis76
WAIT! i think i messed up? on part (1) i used 373-255,
aaronq
  • aaronq
oh damn, you did, i thought it read 273 K
lilsis76
  • lilsis76
haha q1 is 43.02
aaronq
  • aaronq
haha thats more like it 282 kJ is a lot lol
lilsis76
  • lilsis76
okay NOW i have 2.54*7.52(373-273) = 1910.08 right?!
aaronq
  • aaronq
aweseome!
lilsis76
  • lilsis76
haha am i doing the problem right after we fixed it? dang... about 10 more minutes D:
aaronq
  • aaronq
yes, it's correct.. so far :P step 4 !
aaronq
  • aaronq
vaporizing the water
lilsis76
  • lilsis76
haha so far psssh..... haha okay STEP 4! :D
lilsis76
  • lilsis76
okay i get 40.7 kJ/mol (18.02 g h2o)= 733.41 kJ/ g*mol
lilsis76
  • lilsis76
right @aaronq
aaronq
  • aaronq
hmm you need to use moles of water, you wrote the molar mass of water instead
lilsis76
  • lilsis76
dang it so i gotta switch 18 g h20 with....... 2.54 moles right?
aaronq
  • aaronq
yes
lilsis76
  • lilsis76
okay i get 103.378 moles H20
lilsis76
  • lilsis76
now, step 5???? ! :D
aaronq
  • aaronq
103.378 moles H20? how did you get more moles of water? :P
lilsis76
  • lilsis76
shoot hold on
lilsis76
  • lilsis76
40.7 kJ/mol (2.54 mol ) = 103.4 right? what did i do wrong?
aaronq
  • aaronq
103.4 units!!
lilsis76
  • lilsis76
haha help me please what did i do??? i gotta leave in a bit and i think ill be off the computer for the night :/
aaronq
  • aaronq
103.4 \(\Huge{kJ}\)
aaronq
  • aaronq
just watch your units. step 5!
aaronq
  • aaronq
it's very very easy to make mistakes if you don't really pay attention to the units.
lilsis76
  • lilsis76
dang it! i didnt put kJ haha but it was right. okay STEP 5!
aaronq
  • aaronq
okei!
lilsis76
  • lilsis76
2.54mol(.904kJ/mol)(380-373) = 16.073 kJ K RIGHT?!
aaronq
  • aaronq
\(\checkmark\checkmark\checkmark\)
lilsis76
  • lilsis76
okay now i add them all up right and get...
aaronq
  • aaronq
the final answer.
lilsis76
  • lilsis76
2181.031 kJ right???
aaronq
  • aaronq
idk it sounds right. i haven't been checking you arithmetic, i trust you know how to add jk
aaronq
  • aaronq
but if you didn't make mistakes inputting the stuff in your calculator, that should be the right answer.
aaronq
  • aaronq
for the test, you should double check every calculation. It's easy to press a wrong number
lilsis76
  • lilsis76
oh.... okay . welll lol did it look right to you?
aaronq
  • aaronq
yes 2000 kJ sounds about right
lilsis76
  • lilsis76
YAY!!! :D
aaronq
  • aaronq
woo! you should rework this problem again (by yourself) before your test

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