# Thermal Expansion problem

1. Nov 5, 2006

### S_fabris

Here is a two part problem I'm having trouble with:

The length of the column of mercury in a thermometer is 4cm when the thermometer is immersed in ice water and 23.5cm when the thermometer is immersed in boiled water.
a) What should be the length at room temperature 22degreesCelcius?
I did ratio calculation to find this and got 8.23cm and it was correct

Here is where I'm stuck:
b) if the mercury column is 25.2cm long when the thermometer is immersed in a chemical solution, what is the temperature of the solution (answer in deg.Celsius)

Again i tried doing a ratio: 100degC = 23.5cm
xdegC = 25.2cm
my answer was 107.23degC and I is incorrect...though it makes sense and seems correct to me logically...I'm not really sure where to go from here any advice????

Sergio

2. Nov 5, 2006

### ultimateguy

Try converting the degrees to Kelvin and see if that works.

Edit: I seem to recall an expansion formula such as L=Li*alpha*(T2-T1) where L is the initial length and Li is the length by which it changes. Alpha is a constant. Maybe try solving for T2. Don't know if this is right though.

Last edited: Nov 5, 2006
3. Nov 5, 2006

### Max Eilerson

Not sure what you mean by ratio.

23.5cm - 4cm = 19.5cm over 100 C, so each 1 C division is 0.195cm. Go from there

4. Nov 6, 2006

### S_fabris

If I use this information...

0.195cm = 1 C
25.2cm = x C

so 0.195(x) = 25.2(1)
x = 25.2/0.195
x = 129.23 C (incorrect)

that what i meant by "ratio" (i study in French don't know English terms that well :P)

so so far i know that 129.23C and 107.23C are incorrect :S

5. Nov 6, 2006

### Max Eilerson

(25.2cm - 23.5cm) = 1.7cm, you just need to work out how many degrees C 1.7cm represents. You then add this to 100 C ( the temperature at 23.5cm)

6. Nov 6, 2006

### Max Eilerson

Your method is 'wrong' because the x=0cm doesn't represent T = 0 C it represents x= 4cm, T= 0 C.

7. Nov 6, 2006

### S_fabris

Thanks for clearing this up for me Max (and your patience)

Something just wasn't clicking i guess...i got the correct answer at 108.72C :D

Thx again :)