Thermal Expansion of thermometer in ice water

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SUMMARY

The discussion centers on calculating the temperature of a chemical solution using the thermal expansion of mercury in a thermometer. The initial lengths of the mercury column at 0°C (4cm) and 100°C (23.5cm) were provided, leading to a correct calculation of the length at room temperature (22°C) as 8.23cm. However, the incorrect attempts to find the temperature corresponding to a 25.2cm mercury column were clarified through the use of the thermal expansion formula L = Li + α(T2 - T1), ultimately leading to the correct temperature of 108.72°C.

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  • Understanding of thermal expansion principles
  • Familiarity with the mercury thermometer's functioning
  • Knowledge of temperature scales (Celsius and Kelvin)
  • Ability to apply linear equations and ratios in problem-solving
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  • Study the thermal expansion formula L = Li + α(T2 - T1) in detail
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S_fabris
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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?:confused: :confused:

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

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