Question

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.

Answer 1

okay let me try

Answer 2

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

Answer 3

damn, i mean K :S sorry.

do you have the value for the enthalpy of sublimation?

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

okay sorry, i interpreted it wrong, so you need to do this in 4 steps.


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.

Answer 8

i meant 5 steps not 4****

Answer 9

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

Answer 10

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

Answer 11

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)

Answer 12

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

Answer 13

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.

Answer 14

oh.

Answer 15

well then I have nothing else to go by :(

Answer 16

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.

Answer 17

okay im writing now

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

okay, im ready. whats next

Answer 26

okay so step 1) find the heat req'd to raise the temp of the ice to 273 K (zero degrees celsius)

use: \(q=m*C*(T_f-T_i)\)

Answer 27

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

Answer 28

right? it looks right but I think its missing alot

Answer 29

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?

Answer 30

i just have the 255K and 380K

Answer 31

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

Answer 32

look at the graph i drew

Answer 33

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

Answer 34

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

Answer 35

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 \(T_i?\)

Answer 36

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

Answer 37

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?

Answer 38

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

Answer 39

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

Answer 40

ok ok so is it like. this? hold on

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

yep thats right.

Answer 45

YAY! i did 1 thing right! :D

Answer 46

haha woo ! now step 2 !

Answer 47

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 ???

Answer 48

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

Answer 49

so step 2?

Answer 50

sorry said no connection haha

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

\(\checkmark\)