Question

Calculate the enthalpy change, ΔH, for the process in which 14.5g of water is converted from liquid at 10.5∘C to vapor at 25.0∘C .

For water, ΔHvap = 44.0kJ/mol at 25.0∘C and Cs = 4.18 J/(g⋅∘C) for H2O(l).

Answer 1

aaronq you're my hero.

Answer 2

so were assuming that water can vaporize at 25 celsius because the pressure is lower? lol

I think that would change the values you were given. Do you have to account for that?

If not you can just do: \(q=m*C_p*\Delta T+n*\Delta H_{vap}\)

n=moles, btw; you have to use it because the value you were given is in moles.

Answer 3

okay that makes sense because the little sidebar of information includes that formula.

Answer 4

would it be 1 mole?

Answer 5

nope, the molar mass of water is 18g/mol

so just a little less, the formula for moles is: \(n=\dfrac{m}{M}\)

Answer 6

okay so now i'm confused.
What I did was 14.5g * 4.18 J/g-C * 14.5 + n *44.0kj/mol

Answer 7

So for n it would be 18g/1 mol H2O

Answer 8

\(n=\dfrac{m}{M}=\dfrac{14.5~g}{18~g/mol}\)

Answer 9

oh okay
that makes sense.

Answer 10

i got 914.285

Answer 11

but i have a feeling that i'm wrong..

Answer 12

well the answer was 35.4kJ
i had to figure out q for the energy of warming it up and then q for the second half and add the two together.

Answer 13

you didn't take the units into account, the first part is in Joules (look at the units on the specific heat capacity), the second in kJ.

\(1=(14.5* 4.18 * 14.5)J*\dfrac{1kJ}{1000~J} +(\dfrac{14.5}{18} *44.0)kJ\)

Answer 14

sorry that 1 is supposed to be a q

Answer 15

Yeah i realize that now =[

Answer 16

damn, sorry!

Answer 17

it's okay! Thank you for all your help though!