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.

- lilsis76

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- lilsis76

okay let me try

- 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

damn, i mean K :S sorry.
do you have the value for the enthalpy of sublimation?

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- lilsis76

ugh.... no.
this is a practice take home test cuz our exam is tuesday and this is all I have

- 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

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

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

i meant 5 steps not 4****

- 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

haha okay @aaronq let me see well for the
1) 45.7g ( I have to get it to ....kJ or Joules Right?

- 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

so.....45.7g( 1mol/18gh2o) so it would be 2.53 mols? am i in the right direction? @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

oh.

- lilsis76

well then I have nothing else to go by :(

- 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

okay im writing now

- lilsis76

okay im going to search ill be back in a couple mintues

- 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

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

I dont have enthalphy fusion, or enthalpy vaporization in my book :( @aaronq

- aaronq

enthalpy of fusion = ice melt -->6.02 kJ/mol
enthalpy of vaporization = Liquid vaporizing-->40.7 kJ/mol

- aaronq

okay, since all your values are in moles, you do need to convert grams to moles.

- lilsis76

OH 0.o I guess i put it but I didnt recognize it as entalpys

- lilsis76

okay, im ready. whats next

- 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

okay so for number 1:
q= m C A T would be....... 45.7g( 6.02 kJ/mol) ( Tfinal -T initial )

- lilsis76

right? it looks right but I think its missing alot

- 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

i just have the 255K and 380K

- aaronq

but you're only warming up the ice to it's melting point.

- aaronq

look at the graph i drew

- lilsis76

okay so..... i have.. okay so... We have a solid ice at... 255K
a liquid at 273K

- lilsis76

okay @aaronq so were only staying between the temperatures of 255K to373K since the ice is a sold, then its liquid. right?

- 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

Im so lost @aaronq :( I cant figure out how to start it

- 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

i think i have to use the .941kJ/ 1mol and the 45.7 grams right?

- aaronq

yes and no, you have to use moles because the specific heat capacity (0.941 kJ/mol) is in moles.

- lilsis76

ok ok so is it like. this? hold on

- lilsis76

45.7g* \[(\frac{ .941 J }{ 1 mol }) (373k-255k)\]

- 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

how do I do that again? the grams to moles? is that.... 45.7g(1mol/18.02gwater)=2.54moles water?

- aaronq

yep thats right.

- lilsis76

YAY! i did 1 thing right! :D

- aaronq

haha woo ! now step 2 !

- 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

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

so step 2?

- lilsis76

sorry said no connection haha

- lilsis76

okay @aaronq Im calculating and got this..
282.03 is it K or kJ?

- aaronq

K is Kelvin a unit of temperature, kJ is kilojoules (1000 joules). We're finding heat energy here

- lilsis76

okay were finding heat energy. so it would be in kJ so the answer is
282.03 kJ right?

- aaronq

\(\checkmark\)

- lilsis76

YAY!!! I did good for part 1 right? :D

- aaronq

yes! so write down \(q_1=282.03 kJ\) because you're gonna need it at the end.

- aaronq

step 2: \(q_2=\Delta H_{fusion}*n\)

- lilsis76

yes I did okay PART 2! its q= \[\Delta H(m)\]

- lilsis76

is n the number we just got?

- lilsis76

okay @aaronq q=delta H fus (n)
q= 6.02 kJ/mol (282.03 kJ ) right?

- 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

n= moles of H2O

- lilsis76

OH! okay so thats the 18.02g

- 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

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

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

no worries ! lets do this quick then.
yep, part 3!

- lilsis76

okay., part 3 Is 1=mC Delta T

- aaronq

now that you've melted the ice, you gotta raise the temp of the water to it's boiling point (373 K)

- aaronq

yep, but since C is in kJ/mol, you need to substitute m for n
so q=n*C*dT

- aaronq

essentially, what I'm saying is use moles not mass

- lilsis76

okay

- lilsis76

2.54moles(7.52 kJ/mol)( Tfinal-373) what do i put for Tfinal? @aaronq

- aaronq

the temperatures are incorrect,
you melted the ice, right? so that was 273 K
boiling point of water 373 K

- lilsis76

WAIT! i think i messed up? on part (1) i used 373-255,

- aaronq

oh damn, you did, i thought it read 273 K

- lilsis76

haha q1 is 43.02

- aaronq

haha thats more like it 282 kJ is a lot lol

- lilsis76

okay NOW i have 2.54*7.52(373-273) = 1910.08 right?!

- aaronq

aweseome!

- lilsis76

haha am i doing the problem right after we fixed it? dang... about 10 more minutes D:

- aaronq

yes, it's correct.. so far :P
step 4 !

- aaronq

vaporizing the water

- lilsis76

haha so far psssh..... haha okay STEP 4! :D

- lilsis76

okay i get 40.7 kJ/mol (18.02 g h2o)= 733.41 kJ/ g*mol

- lilsis76

right @aaronq

- aaronq

hmm you need to use moles of water, you wrote the molar mass of water instead

- lilsis76

dang it so i gotta switch 18 g h20 with....... 2.54 moles right?

- aaronq

yes

- lilsis76

okay i get 103.378 moles H20

- lilsis76

now, step 5???? ! :D

- aaronq

103.378 moles H20? how did you get more moles of water? :P

- lilsis76

shoot hold on

- lilsis76

40.7 kJ/mol (2.54 mol ) = 103.4 right? what did i do wrong?

- aaronq

103.4 units!!

- 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

103.4 \(\Huge{kJ}\)

- aaronq

just watch your units.
step 5!

- aaronq

it's very very easy to make mistakes if you don't really pay attention to the units.

- lilsis76

dang it! i didnt put kJ haha but it was right. okay STEP 5!

- aaronq

okei!

- lilsis76

2.54mol(.904kJ/mol)(380-373) = 16.073 kJ K RIGHT?!

- aaronq

\(\checkmark\checkmark\checkmark\)

- lilsis76

okay now i add them all up right and get...

- aaronq

the final answer.

- lilsis76

2181.031 kJ right???

- aaronq

idk it sounds right. i haven't been checking you arithmetic, i trust you know how to add jk

- aaronq

but if you didn't make mistakes inputting the stuff in your calculator, that should be the right answer.

- aaronq

for the test, you should double check every calculation. It's easy to press a wrong number

- lilsis76

oh.... okay . welll lol did it look right to you?

- aaronq

yes 2000 kJ sounds about right

- lilsis76

YAY!!! :D

- aaronq

woo! you should rework this problem again (by yourself) before your test

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