# Homework Help: Thermal Expansion of ethanol problem

1. May 26, 2010

### pat666

1. The problem statement, all variables and given/known data
1. You pour 108 cm3 of ethanol, at a temperature of - 10.0 C, into a graduated cylinder initially at 20.0 C, filling it to the very top. The cylinder is made of glass with a specific heat of 840 J/kgK and a coefficient of volume expansion of 1.2 × 10-5 K-1; its mass is 0.110 kg. The mass of the ethanol is 0.0873 kg. The specific heat capacity of ethanol is 2428 J/kgK and the coefficient of volume expansion of ethanol is 75 × 10-5 K-1.

2. Relevant equations

3. The attempt at a solution
V_loss=V_fe-V_fg
=7.677*〖10〗^(-7) m^3
=7.7cm^3
This seems like a very small loss, does anyone know if it correct or not?

2. May 27, 2010

### kuruman

What is the question? What equations did you use and what numbers did you put in them?

V_loss=V_fe-V_fg is not a very informative equation. How did you calculate the numbers that went in it?

3. May 27, 2010

### pat666

yeah i know what you mean.
I just calculated the change in volume for glass (contracted) and then the change in volume for ethanol. From there i found the final volume (once in equilibrium). all using deltaV/V=beta*deltaT.... the equation " V_loss=V_fe-V_fg" i figured out since the etanol was overflowing by the amount it was bigger than the volume of the glass??

4. May 27, 2010

### kuruman

You still don' show how you calculated the changes of each volume. What did you use for the final temperature of each mass and how did you find that? If you don't show us exactly what you did, we will not be able to figure out where you went wrong.

5. May 27, 2010

### EzaMoo

Final temp being -0.89 deg C

Total overflow = 108 x 75 x 10^-5 x (-10+0.89) + 108 x 1.2 x 10^-5 x (20 +0.89)
= 0.765 cm^2

6. May 27, 2010

### kuruman

The calculation is correct, however you should stick to convention and write

ΔT = Tfinal - Tinitial.

It is easier to see that way how ecah volume changes.

7. May 28, 2010