# Homework Help: Maximum Mass You Could Boil with 1000 J of Heat (Thermodynamics)

1. Nov 30, 2013

1. The problem statement, all variables and given/known data
What is the maximum mass of ethyl alcohol you could boil with 1000J of heat, starting from 18∘C?

2. Relevant equations
Q=mcΔT

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

3. 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 :)

2. Nov 30, 2013

### Staff: Mentor

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.

3. Nov 30, 2013

### haruspex

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.

4. Nov 30, 2013

### Staff: Mentor

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.

5. Nov 30, 2013

### SteamKing

Staff Emeritus
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.)

6. Nov 30, 2013

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

Last edited: Nov 30, 2013
7. Dec 4, 2017