Showing posts with label Expansion of Solids. Show all posts
Showing posts with label Expansion of Solids. Show all posts

Expansion of Solids Problems with Solutions Two

We are solving series of problem on the topic expansion of solids. When ever heat energy is given to a solid materiel, the molecules starts vibrating about their mean position and when the given heat energy is withdrawn, they fail to get back to their original positions due to lack of perfect elasticity. Thus there is a dislocation in the positions and that leads to the expansion of the solid material. Its expansion along the length is measured with coefficient of linear expansion, along area is measured with coefficient of areal expansion and along volume is measured with coefficient of volume expansion. We can see lot of real life applications of expansion of solids in daily life. A small gap is left between two rails of railway track so that they can expand in the gap during the summer with out disturbing the structure. Telephone and electric wires on the poles are fixed little loosely so that they can contract in the winter without breaking. Tungsten is used as a filament in electric bulbs as the coefficient of linear expansion is close to the coefficient of linear expansion of glass and with the rise in temperature is same for both of them with the given heat energy.

Problem

Time period of pendulum clock at 25 degree centigrade is given in the problem as two seconds. It is given that at a higher temperature it looses certain time and basing on that we need to find the coefficient of linear expansion of the material and the problem is as shown in the diagram below.


Solution

With the raise of the temperature length of the pendulum increases and hence its time period also increases. Thus the pendulum makes less oscillations per day and hence there is a loss of the time with the raise in the temperature. We have derived a formula for that and by using it we can solve the problem as shown in the diagram below.


Problem

A steel rod of known area of cross section is tightly fixed between two rigid supports and it is not allowed to expand even when its temperature is raised and we need to find the thermal stress developed in the rod. Problem is as shown in the diagram below.


Solution

With the given heat energy the body is supposed to expand but we have fixed it tight such that it cannot expand. Hence a tension and stress is developed in the rod and we can express it basing on the young’s modulus of the rod. In the place of expansion, we shall substitute the value basing on the coefficient of linear expansion definition and the problem can be solved as shown in the diagram below.


Problem

A brass rod and steel rod are having different lengths at a given temperature. We need to find out at what common temperature their final lengths will be equal. Problem is as shown in the diagram below.


Solution

We know that as their coefficient of linear expansion are different, they are going to expand differently with the raise in the temperature. By writing their final lengths equation in both the cases and by equating them we can solve the problem as shown in the diagram below. Coefficient of linear expansion of the materials is given to us. We can solve the problem as shown in the diagram below.


Problem

A cube of known side and coefficient of linear expansion is given in the problem. We need to find the increase in its surface area with the raise of the given temperature and the problem is as shown in the diagram below.


Solution


We know that change in the area is measured in terms of coefficient of areal expansion and it is double to that coefficient of linear expansion. By writing its definition, we can solve the problem as shown in the diagram below. 


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Temperature and its Measurement 

Expansion of Solids Problems with Solutions One

We are solving series of problems on the topic called expansion of solids. When heat energy is given to a solid material, its molecules starts vibrating about their mean position and when the heat energy is withdrawn they try to come back to their original position due to their restoring forces among the molecules. But because none is perfectly elastic, they fail to come back to their original positions and hence there is change in the shape and it is called expansion. This can happen along the length called linear expansion, along area called areal expansion and can happen along the volume called volume expansion. To measure this expansions, coefficient of linear expansions are defined. We can also find the relation between them and found that they are respectively in the ratio of 1:2:3.

Problem

Coefficients of linear expansion of two different metals are in the ratio of 3:4. We need to know the ratio of initial lengths so that for the same rise in temperature the expansions will be the same and the problem is as shown in the diagram below.


Solution

We know that the coefficient of linear expansion is defined as the ratio of increase in the length of the rod to its original length per unit rise in the temperature. Basing on that we can write increase in the length of the rod as the product of initial length,coefficient of linear expansion and the change in the temperature. As rise in temperature and and increase in the length is same, the product initial length and coefficient of linear expansion is constant. Thus initial lengths ratio is reciprocal to the coefficients ratio and the solution is given as shown in the diagram below.


Problem

A brass disc is having known diameter and it is having a hole of known diameter raised to a certain temperature and we need to know the increase in the size of the hole and the problem is as shown in the diagram below.


The hole par t of disc expands as if there is content there and we need to use the basic definition of linear expansion to solve the problem as shown in the diagram below.


Problem

Two different rods are having two different coefficient of linear expansions and at all temperatures their difference in terms of lengths remains constant and we need to find the initial lengths of both the rods.Problem is as shown in the diagram below.


Solution

Their initial lengths are different and hence there is a difference of lengths between them. When we raise the temperature both the rods expands and it is going to be different. For the difference between the lengths remain constant, the increase in the length of each rod has to be same for the given temperature. Taking this as the condition, we can solve the problem as shown in the diagram below.


Problem

A steel rod of half kilometer is used in the construction of the bridge and it can withstand a maximum temperature of 40 degree centigrade. Coefficient of linear expansion is given to us and we need to find the gap that has to be left so that it can fit with out causing any problem. Problem is as shown in the diagram below.


Solution

We know that the gap that has to be left is the size to which the rod can expand. We can measure the increase in the size of the rod as we know the initial length, rise in temperature and coefficient of linear expansion. Problem is solved as shown in the diagram below.




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Temperature and its Measurement 

Thermal Properties of Heat Complete Lesson

Heat is a disordered form of energy and to know the direction of flow of the heat, we need the concept of temperature. Temperature is a measure of the heat energy. Heat is measured with a unit called calorie and temperature is measured with kelvin in the standard international system. The rise in the temperature of a body causes the expansion in general in the body and it is called thermal expansion. To study this expansions, we need coefficients of expansions and they were separately defined for solid state, liquid state and gaseous state materials. 

Heat flows from one place to other in different ways called conduction, convection and radiation. Conduction need a solid medium, convection need a fluid media and radiation is not in need of any media for the propagation.

Here in this lesson we have discussed about all this topics in detail.

Expansion of Solids and Applications
                                                     

Anomalous expansion of water 

Expansion of Liquids Problems with Solutions


Heat and Thermodynamics Complete

Heat is a form of energy. Heat can be converted to other forms of energy’s and other forms of energy’s can also be converted into heat. Heat can be measured with a physical quantity called temperature. When heat energy is given to your body it expands. The expansion has to be studied separately for solids, liquids and gases.

The conversion of the heat energy into other forms and its applications are studied in the chapters called calorimetry and thermodynamics. The reason behind the temperature is nothing but the internal collisions of molecules of a gas and it is studied in the kinetic theory of gases.

This post is a collection of all the topics that are relevant to the heat and Thermo dynamics.

Posts available in the blog are 


Problems on Expansion of Solids

Problem and solution

In a iron disc there is a hole at the center. A brass rod of equal size is exactly fixed in the hole. To get the brass rod out of the hole, what shall we do with respect to temperature?

To solve this problem we shall understand one simple concept. Coefficient of linear expansion is a measure of expansion of the material with respect to the rise of the temperature. It is also a measure of contraction of a material with respect to the decrease in temperature.

If a material is having a higher value of coefficient of linear expansion, with the increase of the temperature at expands more and with the decrease of the temperature it contracts more.

We know that coefficient of linear expansion of brass is more than that of iron. Hence if you rise the temperature brass expands more and it becomes further very tight in the hole.

If you decrease the temperature brass contracts more therefore it can be taken out from the hole quite easily.



Problem and solution

Two different materials are having different lengths at a certain temperature. What is the condition for the difference between the two material lengths always remains constant at all temperatures?

If the difference between the two lengths has to remain constant, both of them shall increase their lengths by the same value with respect to the increase in the temperature. So we shall equate the increase in the lengths of both of them.



Problem and solution

Find the ratio of expansions in the two wires when same heat energy is supplied to both of them?

When same temperature is given to both of them we can say that the increase in the length of the materials is directly proportional to its length, coefficient of linear expansion and the rise in the temperature.

When the same heat energy is given to both of them they are not going to expand equally. The material with more specific heat will expand less and vice versa. So we shall take the concept of specific heat into consideration while solving this problem. It can be proved that the increase in the length in this case is going to be independent of the length of the material but depends on the area of its cross-section.




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Expansion of Solids and Applications

We know that matter exists in three different states. They are solid, liquid and gaseous states. Whenever heat energy is supplied to any of the states there is a increase in the size and it is called expansion. The basic reason for the expansion is simple. The different states of the matter will have molecules and with a certain separation between them. When the temperature is increased and there is a permanent increase in the distance between the molecules which leads to the expansion.

In the case of the solid material the force of attraction between the molecules is very strong and the molecules are very tightly packed with each other. When the heat energy is given the molecules absorb the heat energy and starts vibrating about the mean position. When the heat energy is withdrawn the molecules are supposed to come back to their original position and a body is also supposed to get its original state. But due to the non-harmonic nature of the molecule vibration, there is a permanent increase in the distance between the molecules. Because of lack of perfect elasticity, the molecules are unable to come back to their original positions which lead to a permanent expansion.

A solid material can expand along its length, area and along its volume. To measure the expansion along the length, we have coefficient of linear expansion. To measure the expansion along the area, we have coefficient of year expansion. To measure the expansion along the volume, we have coefficient of volume expansion.

Coefficient of linear expansion is defined as the ratio of increase the length of the material to its original length the per 1°C rise in temperature. It can be measured with the unit of 1°C or per degree Kelvin.

Coefficient of areal expansion is defined as the ratio of increase in the area of the material to its original area per 1°C rise in temperature.

Coefficient of volume expansion is defined as the ratio of increase in the volume of the material to its original volume per 1°C rise in temperature.

All these coefficients of expansion depend on the nature of the material but is independent of physical dimensions of the body. It can be clearly notice that coefficient of volume expansion is three times the coefficient of areal expansion is  two times that of the coefficient of linear expansion.

We can derive the relation between them in a simple format as shown below. To derive the relation we have considered a body in a cube shape with the unit dimensions. Let us rise the temperature of this body by only 1°C and we can write the equation for the final length, final area and the final volume as shown.




We have also derived the equation further ratio of change in the moment of inertia to its original moment of inertia in terms of coefficient of linear expansion in the above equation. During this derivation we have assumed that the change in the radius is small and hence we have applied the concept of approximation.

 Applications of expansion of solids

Expansion of the solids is used in our daily life that so many occasions.

1. For example telephone and electric wires are arranged little bit loosely between any of the two poles. It is simply because with the decrease in temperature in winter season, the wire contracts and if they were arranged tightly that will become further tight and they may break. To avoid this problem, they were arranged loosely.

2.Between the two rails on a railway track always a small gap is left. In the summer with the increase in temperature they expands as this materials are made up of metals. One end of the rail is fixed and other end is having a small gap with the next one. The expansion can happen in the gap therefore the track will remain intact.

3. In electric glass bulbs tungsten is used as a filament. The linear expansion coefficient of the glass is close to that of the tungsten therefore with the increase in the temperature both of them expands equally so that the use is not going to fail for a long time.

4. When water is sprinkled on a hot chimney glass only at the particular points the glass contracts and that is the reason why the glasses going to break.

The loss or gain of a time with the pendulum clock

1.The pendulum clock consists of the pendulum which is made up of a metal.

2.In the summer when temperature is raised the length of the pendulum increases and hence it’s time period also increases.

3.It means it is going to take more time to make one oscillation.

4.It implies it is going to make less number of oscillations in a given time. It means it is going to show less time than required when the temperature is raised.

5. In the winter season the temperature decreases and hence the pendulum makes more number of oscillations than required and hence it shows again of time.

6. We can measure the loss or gain of the time using a mathematical equation. We can write the equation for the time period of a pendulum using a concept that time period is directly proportional to Squire route of its length.

7. The equation further difference in the time with respect to the variation of the temperature is as shown below.




In the place of the time period we can write the number of the seconds per day when we are calculating the loss/gain of the time with respect to the pendulum per day. Depending on the question we can write any number of the seconds as per the demand of the problem. For example if we have to calculate the difference in the time for half today we have to write the number of the seconds per half-day only.

Thermal stress

Assume that a wire of certain modulus of elasticity is arranged between two rigid supports. When the temperature of the system is reduced the wire tries to contract. As it is permanently fixed, it cannot contract. As a result stress is developed on the wire and the stress and is called thermal stress. In the above diagram we have derived the equation for this thermal stress also.

Problem and solution

A pendulum clock loses 10 seconds per day at 40°C and gains five seconds per day at 20°C. At what temperature the pendulum clock shows correct time?

We know it very clearly that pendulum clock shows less time with a higher temperature and more time at a lower temperature.

It implies that the pendulum clock shows the correct time at a temperature less than 40°C and  more than 20°C.

We can write the equation for the loss of the time in the first case another gain of the time in the second case as shown below.



We can design a pendulum with a bimetallic strip so that it can always show correct time.

The pendulum shall consists of two different materials of similar sizes arranged together.

One metal with the increase of the temperature shall be arranged in such a way that it moves in the upward direction.

Simultaneously the other metal shall move in the downward direction.

 We shall arrange such that the expansion of the first metal in the upward direction so the length of the pendulum increases.

And the expansion of the second material shall happen in downward direction the pendulum length decreases.

If they are arranged in such a way that the increase in the length of both the metals is same then the length of the pendulum comes back to its original value and hence it always shows the correct time.



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