Linear and Volume Expansion problem

In summary, the volume of glycerine in an aluminum container at a temperature of 105.2°C can be calculated by first finding the volume of glycerine at 20°C, which is 462.7 mL. Then, using the volume expansion coefficient of glycerine (4.85 × 10^-4 (°C)-1) and the linear expansion coefficient of aluminum (2.40 × 10^-5 (°C)-1), the volume of glycerine at 105.2°C can be calculated to be 481.8 mL. However, due to the expansion of the aluminum container, the scale reading will be smaller, at 478.9 mL, which can be found by multiplying
  • #1
MMONISM
20
1

Homework Statement


An aluminium container (cylinder) is filled with glycerine. Glycerine has a volume expansion coefficient of 4.85 × 10^-4 (°C)-1. Aluminium has a linear expansion coefficient of 2.40 x 10^-5 (°C)-1. At a temperature of 20.00°C the scale on the side of the aluminium container reads that there are 462.7 mL of glycerine in the container. At 105.2°C what volume of glycerine will the scale read?

Homework Equations

[/B]
[tex]\Delta L = \alpha L \Delta T[/tex]
[tex]\Delta V = \beta V \Delta T[/tex]

The Attempt at a Solution


When the aluminium container is 20 degree, 1 ml is 1 ml in scale, when it is 105.2 degree, the scale increase by 2.40 x 10^-5 x 1 x (105.2-20) =0.0020448 ml. Then I am not sure what the next step is.
 
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  • #2
The next step is to correct your previous step. How is β of a material related to α of the same material?

Chet
 
  • #3
Chestermiller said:
The next step is to correct your previous step. How is β of a material related to α of the same material?

Chet
β=3α? is this also work for liquid, is this mean I need to divide β by 3 to get the linear expansion for glycerine?
 
  • #4
MMONISM said:
β=3α? is this also work for liquid, is this mean I need to divide β by 3 to get the linear expansion for glycerine?
No. You need to multiply the α of aluminum by 3 to get the volume expansion of the aluminum.

Chet
 
  • #5
I will try again, Thanks
 
  • #6
Chestermiller said:
No. You need to multiply the α of aluminum by 3 to get the volume expansion of the aluminum.

Chet

Could you tell if I am correct please?

αa = 2.4*10^-5 (°C)-1, βa = 3*αa = 7.2*10^-5 (°C)-1. therefore 1 ml in 20.00°C will become 1 * 7.2*10^-5 * (105.2 - 20) + 1 = 1.00613 ml.
which means if put 20.00°C,462.7 mL( in 20.00°C aluminium container) glycerineat into 105.2°C aluminium container, the reading will be smaller which is 462.7/1.00613 = 459.88 mL.
Then if 20.00°C,459.88 mL glycerine (in 105.2°C aluminium container) increase temperature to 105.2°C the final reading would be = 459.88 * 4.85 × 10^-4 * (105.2-20) + 459.88 = 478.88 mL
 
  • #7
MMONISM said:
Could you tell if I am correct please?

αa = 2.4*10^-5 (°C)-1, βa = 3*αa = 7.2*10^-5 (°C)-1. therefore 1 ml in 20.00°C will become 1 * 7.2*10^-5 * (105.2 - 20) + 1 = 1.00613 ml.
which means if put 20.00°C,462.7 mL( in 20.00°C aluminium container) glycerineat into 105.2°C aluminium container, the reading will be smaller which is 462.7/1.00613 = 459.88 mL.
Then if 20.00°C,459.88 mL glycerine (in 105.2°C aluminium container) increase temperature to 105.2°C the final reading would be = 459.88 * 4.85 × 10^-4 * (105.2-20) + 459.88 = 478.88 mL
I don't think this is correct.

If the original volume at 20 C is 462.7 mL , what will be the actual volume of the aluminum container at this marker after the container has been heated to 105.2 C? (The open volume has to increase, not decrease).

The volume of glycerine at 105.2 C is calculated correctly.

Chet
 
  • #8
Chestermiller said:
I don't think this is correct.

If the original volume at 20 C is 462.7 mL , what will be the actual volume of the aluminum container at this marker after the container has been heated to 105.2 C? (The open volume has to increase, not decrease).

The volume of glycerine at 105.2 C is calculated correctly.

Chet

Sorry, I think I didn't make myself clear enough, of course if the container increase temperature it's open volume will increase, but in the first paragraph, I actually try to say that after container increase temperature, if we put 462.7 mL glycerin in 20 C scale to 105.2 C scale the reading will be smaller, as temperature of the liquid is still 20 C, volume doesn't change and therefore if moving it to a bigger scale, the reading is smaller. Is this correct?

If my calculation for glycerin at 105.2 C is correct, is this the final reading from the container at 105.2 C? Thanks
 
  • #9
MMONISM said:
Sorry, I think I didn't make myself clear enough, of course if the container increase temperature it's open volume will increase, but in the first paragraph, I actually try to say that after container increase temperature, if we put 462.7 mL glycerin in 20 C scale to 105.2 C scale the reading will be smaller, as temperature of the liquid is still 20 C, volume doesn't change and therefore if moving it to a bigger scale, the reading is smaller. Is this correct?

If my calculation for glycerin at 105.2 C is correct, is this the final reading from the container at 105.2 C? Thanks
I think what you did in post # 6 is correct. I did it a little differently. Here was my approach:

I said that the volume of glycerine increased to 462.7(1+4.85 x 10-4 x 85.2) = 481.8 ml

The volume of the container, up to the 462.7 mark on the scale, increases to 462.7(1+7.2 x 10-5 x 85.2) = 465.5 ml
So, at 105.2 C, the ratio of volume indicated on the scale in the aluminum container to actual volume is (462.7)/(465.5) = 0.994
So the final volume of glycerine indicated on the scale is 0.994 x 481.8 = 478.9 ml. This matches with your value.

Chet
 
  • #10
Chestermiller said:
I think what you did in post # 6 is correct. I did it a little differently. Here was my approach:

I said that the volume of glycerine increased to 462.7(1+4.85 x 10-4 x 85.2) = 481.8 ml

The volume of the container, up to the 462.7 mark on the scale, increases to 462.7(1+7.2 x 10-5 x 85.2) = 465.5 ml
So, at 105.2 C, the ratio of volume indicated on the scale in the aluminum container to actual volume is (462.7)/(465.5) = 0.994
So the final volume of glycerine indicated on the scale is 0.994 x 481.8 = 478.9 ml. This matches with your value.

Chet

Thank you so much, really appreciate you answered all my question with patience !
 

Related to Linear and Volume Expansion problem

1. What is linear expansion?

Linear expansion refers to the increase in length of an object when it is heated. This occurs because the molecules in the object gain energy and vibrate more, causing the object to expand. The amount of expansion is directly proportional to the change in temperature and the original length of the object.

2. What is volume expansion?

Volume expansion is the increase in volume of an object when it is heated. This occurs because the molecules in the object move further apart, taking up more space and causing the object to expand. The amount of expansion is directly proportional to the change in temperature and the original volume of the object.

3. What is the coefficient of linear expansion?

The coefficient of linear expansion is a constant that represents the change in length per unit length per degree of temperature change. It is denoted by the symbol α and is specific to each material. It is used in calculations to determine the amount of linear expansion of an object.

4. How is the change in length or volume calculated for a given material?

The change in length or volume can be calculated by using the formula ΔL = αLΔT or ΔV = βVΔT, where ΔL or ΔV is the change in length or volume, α or β is the coefficient of linear or volume expansion, L or V is the original length or volume, and ΔT is the change in temperature.

5. What are some real-life applications of linear and volume expansion?

Linear and volume expansion are important in various fields such as engineering, construction, and thermodynamics. They are used in the design of bridges, buildings, and pipelines to account for the expansion and contraction of materials due to temperature changes. They are also used in the production of thermometers and other temperature measuring devices. Additionally, the expansion of liquids and gases is utilized in the functioning of engines and refrigeration systems.

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