Maximum Mass You Could Boil with 1000 J of Heat (Thermodynamics)

[adamwest]

Homework Statement

What is the maximum mass of ethyl alcohol you could boil with 1000J of heat, starting from 18∘C?

Homework Equations

Q=mcΔT

c ethyl alcohol = 2400 J/(kg*K)
T boil ethyl alcohol = 78 C = 351 K
T starting = 18 C = 291 K

The Attempt at a Solution

m=Q/(c*ΔT)

m= (2400 J)/[(2400 J/kg*K)(351 K -291 K)] = .00694 kg (or 6.94E-3 kg)

It tells me this is wrong. I am probably overlooking something quite obvious. Thanks :)

 

[Borek]

So far you calculates mass than can be heated to the boiling point. It has not even started to boil, not to mention boiling away.

 

[haruspex]

So far you calculates mass than can be heated to the boiling point. It has not even started to boil, not to mention boiling away.

Well, the OP doesn't say 'boil away'. When I boil water, I bring it to the boil - I don't keep boiling until there's none left. Maybe it's not been quoted correctly.

 

[Borek]

My first idea was it means "boil away", as opposed to "bring to boil". But after your comment I see the wording as ambiguous.

Can be my English fails me and I am completely off.

 

[SteamKing]

Just a nit pick: a temperature difference of 1 degree centigrade is the same temperature difference of 1 degree Kelvin. (You don't need to convert Celsius temps. to Kelvin temps. if you are interested only in temperature differences.)

 

[adamwest]

Borek, you were right (and your English is just fine! :) ). Here is how you get the right answer:

L vaporization ethyl = 8.79 x 10^5 J/kg

Qtotal = Q1 + Q2 = (Energy to heat mass up to boiling point) + (Energy used to boil all of mass)

1000 J = mcΔT + mLv

1000 J = m(cΔT + Lv)

m = 1000 J/(cΔT + Lv)

m = 1000 J/[(2400 J/kg*K)(351 K - 291 K) + 8.79 x 10^5 J/kg)]

m = 9.78 x 10^-4 kg = .978 g

 

[Chad Jensen]

This drove me nuts as well. I read it as how much can you bring to boiling point. Not how much can you vaporize. I am glad I found this.