Superficial dilation: formula, experiment, exercises

Dilationshallow is the name given to the phenomenon in which there is a increase in the area of ​​abody caused by an increase in temperature. This type of expansion occurs in bodies with superficial symmetry, such as plates, table tops, boards, tiles, etc.

Lookalso: Calorimetry

The surface dilation depends on the surface expansion coefficient. This coefficient, whose unit is the °C-1, is feature of each type of material, but keep a proportional relationship with the coefficient of linear expansion:


β — coefficient of surface expansion (°C-1)

α — linear expansion coefficient (°C-1)

We can understand this relationship if we see that in surface dilation there are two dilationslinear: one for the length and another for the height of the body. It is important to note that the relationship shown above is valid only for bodies formed by pure substances and homogeneous.

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Formula

Check the formula that is used to calculate the modulus of surface dilation — the variation in the area suffered by some body when it is heated.


S
— area dilation (m²)

s0— initial area (m²)

β — coefficient of surface expansion (°C-1)

ΔT —temperature variation (°C)

In addition to this way, we can calculate the surface dilation in another way, so that we can directly find the final area of ​​the body, check:


s
— final area (m²)

Thermal expansion

When heated, the molecules of bodies tend to vibrate at higher speeds, this makes the macroscopic dimensions of bodies can be changed, albeit minimally. The phenomenon in which a body changes size when heated is called dilationthermal.

Despite being intuitive, it is not true that all materials expand when their temperature is increased, there are materials that present negative expansion coefficients (such as vulcanized rubber), that is, when these materials are heated, their dimensions are reduced.

Thermal expansion is divided into three expansion subtypes: linear, shallow and volumetric. These types of dilation occur together, however, one of them will be more significant than the others, according to the shape of the body.

For example: due to its shape, the needle suffers more dilationlinear in relation to other forms of dilation; a metal plate, in turn, suffers more dilationsuperficial, because of its format; liquids and gases, which occupy the space of their containers, tend to expand in all directions and, therefore, present dilationvolumetric.

Lookalso:What is entropy?

Dilation of liquids

Liquids can undergo volumetric expansion when heated. However, when studying this type of dilation, it is important that we consider the volumetric expansion of containers where liquids are stored.

In this sense, one speaks of apparent dilation — the difference between the dilation suffered by the liquid and by its container. Access our article and learn all about liquid dilation.

Experiment

There are experiments that can be done quickly and easily, in order to visualize the phenomenon of superficial dilation. Check out some cases:

Necessary materials:

  • 1 Styrofoam tray

  • 1 coin

  • 1 candle

  • matches

  • 1 pliers

  • 1 pen

  • 1 scissors

Methodology:

Place the coin on the Styrofoam tray and outline it with the pen. After that, cut it out. Light the candle and hold the coin with the pliers, positioning it just above the candle flame (be in the presence of an adult to perform this type of experiment).

After a few minutes, place the coin on the Styrofoam tray and you will notice that it will have shrunk in size after melting the Styrofoam. To compare the sizes of the heated and cold coins, place the hole produced by the heated coin and the piece of Styrofoam that was cut out side by side.

  • Another interesting experiment is to have a rim and a metallic sphere with a radius a little larger than the rim. At room temperature, the sphere will not be able to pass through the rim, however, when we heat the rim, its internal area increases due to thermal expansion and the sphere will be able to pass through it:

  • Another possibility is to try to open a pot that has the lid attached by heating it, making its area expand:

solved exercises

Question 1) A 0.05 m² rectangular metal sheet has a temperature of 25º C when it is heated by sunlight, until its temperature reaches 75º C. The coefficient of surface expansion of the material that makes up the sheet being equal to 2.0.10-4 ºC-1, how much will be the variation in the area of ​​this plate?

a) 0.0575 m²

b) 0.0505 m²

c) 1,500 m²

d) 0.750 m²

e) 0.550 m²

Template: Letter B

Resolution:

To find the final area of ​​the sheet metal, we will use the following surface expansion formula:

We will insert, in the formula, the data that were provided in the exercise:

According to the data provided by the exercise, the final area of ​​this metal sheet will be 0.505 m², so the correct alternative is the letter B.

Question 2) A given material has a coefficient of linear expansion of 1.5.10-5 °C-1, the coefficient of surface expansion of this same material is:

a) 0.50.10-5 °C-1

b) 0.75.10-5 °C-1

c) 3.0.10-5 °C-1

d) 4.50.10-5 °C-1

e) 0.40.10-5 °C-1

Template: Letter Ç

Resolution:

To solve this exercise, just remember that two bodies with different symmetries, but made of the same pure substance, keep the following relationship between their coefficients of thermal expansion:

Therefore, the correct alternative is the letter Ç.

Question 3) A 0.4 m² board and surface expansion coefficient equal to 2.0.10-5 °C-1 is heated from 20º C to 200º C. Determine the percent increase in area for that plate.

a) 0.36%

b) 35%

c) 25%

d) 0.25%

e) 5%

Template: Letter THE

Resolution:

Let's first calculate the modulus of the expansion suffered by the plate using the surface expansion formula:

Using the data provided by the exercise, we will have to do the following calculation:

In this resolution, we first calculate what was the dilation suffered by the plaque. Next, we made the ratio between the final area of ​​the plate, which is the sum of the initial area with the expansion of the plate, by the initial area of ​​the plate. After multiplying the value obtained by 100, we find the percentage of the new area in relation to the previous one: 100.036, that is: the area of ​​the plate increased by 0.36%.
By Me. Rafael Helerbrock

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