Thermal Expansion of a Liquid

In summary: The question should have specified that βΔT is small. You are expected to use the first order approximation.Thanks all the same.
  • #1
108
5
Homework Statement
A liquid has a density ##\rho##.
Show that the fractional change in density for a change in temperature ##\Delta##T is
##\frac {\Delta \rho} {\rho} = -\beta \Delta T##
Relevant Equations
##\Delta V = \beta V \Delta T##
##\Delta V = \beta V \Delta T####\rho _2 = \frac m {V + \Delta V} = \frac { \rho V} {V(1+\beta \Delta T)} = \frac \rho {1+\beta \Delta T}##
##\Delta \rho = \rho _2 - \rho = \frac \rho {1+\beta \Delta T} - \rho = \rho (\frac 1 {1+\beta \Delta T} - 1) = \rho \frac {-\beta \Delta T} {1+\beta \Delta T}##

##\frac {\Delta \rho} {\rho} = \frac {-\beta \Delta T} {1+\beta \Delta T}##

I can't figure out were I make the mistake...
It should be:
##\frac {\Delta \rho} {\rho} = -\beta \Delta T##
 
Physics news on Phys.org
  • #2
EEristavi said:
Homework Statement: A liquid has a density ##\rho##.
Show that the fractional change in density for a change in temperature ##\Delta##T is
##\frac {\Delta \rho} {\rho} = -\beta \Delta T##
Homework Equations: ##\Delta V = \beta V \Delta T##

##\Delta V = \beta V \Delta T####\rho _2 = \frac m {V + \Delta V} = \frac { \rho V} {V(1+\beta \Delta T)} = \frac \rho {1+\beta \Delta T}##
##\Delta \rho = \rho _2 - \rho = \frac \rho {1+\beta \Delta T} - \rho = \rho (\frac 1 {1+\beta \Delta T} - 1) = \rho \frac {-\beta \Delta T} {1+\beta \Delta T}##

##\frac {\Delta \rho} {\rho} = \frac {-\beta \Delta T} {1+\beta \Delta T}##

I can't figure out were I make the mistake...
It should be:
##\frac {\Delta \rho} {\rho} = -\beta \Delta T##
There is no mistake, you just need to take the first order approximation.
The ΔT in the denominator is insignificant compared with the 1 in the denominator.
 
  • Like
Likes EEristavi
  • #3
haruspex said:
There is no mistake, you just need to take the first order approximation.
The ΔT in the denominator is insignificant compared with the 1 in the denominator.

##\Delta T## or ##\beta? ##
Because as I think ##\Delta T## Can me "big"

Also,
Can you give me some direction: where I can look it up - Why I can discard that value in denominator?
I know I will find it eventually, but I would appreciate it if you could help me spare some time

P.S. I kind of tried to expand this as a Tailor (maclaurin) Series, but was't satisfied with the answer..
 
  • #4
EEristavi said:
Because as I think ΔT Can me "big"
It can be, but the expression you are asked to derive for change in density is not valid for large changes in temperature.
E.g. consider a temperature gain so great that the volume doubles. βΔT=1. What does that give for the new density, according to the target formula?
 
Last edited:
  • Like
Likes TSny and DEvens
  • #5
I think an easy approach is to simply differentiate the fundamental relationship between volume, mass and density.

##m = \rho V\tag{1} \rightarrow V = m \rho^{-1}##

If we differentiate this for a unit of mass:

##dV = -m\rho^{-2} d \rho \tag{2}##

Now simply plug (1) and (2) into the relation

##d V = \beta V d T \tag{3}##

Your desired equation immediately results.
 
  • Like
Likes EEristavi
  • #6
haruspex said:
What does that give for the new density, according to the target formula?

It would be impossible to use first order approximation - so answer will be "my formula"?
 
  • #7
EEristavi said:
It would be impossible to use first order approximation - so answer will be "my formula"?
The question should have specified that βΔT is small. You are expected to use the first order approximation.
 
  • #8
Thanks all
 

What is thermal expansion of a liquid?

Thermal expansion of a liquid refers to the increase in volume of a liquid due to an increase in its temperature. This is a result of the increase in the kinetic energy of the molecules, causing them to move further apart and take up more space.

What factors affect the thermal expansion of a liquid?

The thermal expansion of a liquid is affected by the type of liquid, its initial temperature, and the change in temperature. Different liquids have different coefficients of thermal expansion, meaning they expand at different rates. The higher the initial temperature of the liquid, the greater the expansion. And the greater the change in temperature, the greater the expansion.

What is the coefficient of thermal expansion?

The coefficient of thermal expansion is a measure of how much a material expands or contracts when its temperature changes. It is defined as the fractional change in length or volume per degree of temperature change.

Why is thermal expansion of a liquid important?

Thermal expansion of a liquid is important in various industries, such as manufacturing, construction, and engineering. It must be taken into consideration when designing structures or machinery to ensure they can withstand the changes in temperature without breaking or malfunctioning.

How is thermal expansion of a liquid measured?

The thermal expansion of a liquid is typically measured using a dilatometer, which is a device that measures changes in volume or length due to temperature changes. The liquid is placed in a sealed container and heated or cooled, and the resulting change in volume is measured and used to calculate the coefficient of thermal expansion.

Suggested for: Thermal Expansion of a Liquid

Replies
2
Views
980
Replies
1
Views
484
Replies
2
Views
889
Replies
10
Views
250
Replies
14
Views
374
Replies
1
Views
714
Replies
3
Views
724
Replies
31
Views
478
Back
Top