Showing posts with label Temperature. Show all posts
Showing posts with label Temperature. Show all posts

Thermodynamics Problems with Solutions Five

We are solving series of problems based on the concepts of thermodynamics. We also deal about calorimetry in this chapter which deals about conversion of heat energy to other forms of energies and its applications further. We do define terms like specific heat and latent heat to explain this properties and they are the basic terms of heat concepts. When there is a change in the temperature, we need to deal with specific heat concept and when there is a change of state, we need to study it in terms of latent heat and during this process, all the supplied heat energy is used to change the state of the system and hence its temperature remains constant.

Problem

During an adiabatic process,pressure of the gas is proportional to the cube of the temperature and basing on that we need to find the ratio of specific heats of the gas. Problem is as shown in the diagram below.


Solution

We need to take the relation between pressure and temperature and taking that into consideration with the given data, we can get the relation between pressure and temperature in the adiabatic process and the problem can be solved as shown in the diagram below.


Problem

A metal sphere of known radius and specific heat is given to us and it is rotating about its own axis with certain rotations per second. When it is stopped half of its energy is converted into heat and we need to measure the raise in the temperature of the system and the problem is as shown in the diagram below.


Solution

As the body is rotating it has rotational kinetic energy and half of it is converted into heat energy as per the given problem. Taking law of conservation of the energy, we can solve the problem as shown in the diagram below.



Problem

The relation between internal energy,pressure and volume is given to us as shown in the diagram below. We need to find the ratio of specific heats and some constants are also available in the problem.


Solution

We need to differentiate the given equation to get the change in internal energy and hence it can be expressed in terms of specific heat of the gas at constant volume. Problem can be further solved as shown in the diagram below.


Problem

Work done by a system under isothermal conditions has to be determined that change its volume from one to other and satisfy the given equation. Problem is as shown in the diagram below.


Solution

Relation of pressure with other physical quantities is given to us as shown in the diagram below. We need to measure the work done and the pressure is not constant here. So to get the work done we shall integrate the pressure with the change in the volume as shown in the diagram below. By simplifying the equation further, we can get the solution as shown in the diagram below.


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Thermodynamics Problems with Solutions Four

We are solving series of problems based on the concept of thermodynamics. It is a branch of physics that deals with the heat energy conversion into other forms and its applications. Carnot engine is a ideal heat engine that converts the given heat into work under a cyclic process. As the process is cyclic process, internal energy remains same and all the supplied heat is going to be converted into work. But we know that heat is a disordered format of energy and it cannot be completely converted into work. That is the reason why, heat engine cannot have hundred percent efficiency. Heat engine will have three basic parts by name source, working substance and sink. Source do supply heat, working substance do convert the heat into work and the extra generated heat to the sink.

Problem

Pressure and volume graph for two different gases during adiabatic process is given to us as shown in the diagram below. We need to know which graph belongs to which gas.


Solution

From the pressure and volume graph, it can be found that the slope of the graph is directly proportional to the ratio of specific heats of the gas. It is clear from the graph that the slope of the second curve is more and hence it shall be the gas with more ratio of specific heat value. Solution of the problem is as shown in the diagram below.


Problem

A refrigerator is placed in a room of temperature of 300 kelvin and the system temperature is 264 kelvin. We need to measure the how many calories of heat shall be delivered to the room to the room for each kilo kelvin of energy consumed by refrigerator by the system ideally. Problem is as shown in the diagram below.


Solution

We know that proficiency of a refrigerator is the amount of work done when compared with the heat energy supplied to the system. It can also be expressed in terms of temperature as shown in the diagram and is solved as shown in the diagram below.


Problem

In carnot’s engine efficiency is 40 % for a certain temperature of the hot source. To increase the efficiency of the system by 50 %, what shall be the source temperature and the problem is as shown in the diagram below.


Solution

We know that the efficiency of heat engine is the magnitude of the work done when compared with the heat energy supplied and it can be expressed in terms of absolute temperature as shown in the diagram below.

We need to apply it for two cases and solve the problem as shown in the diagram below.


Problem

The relation between temperature and pressure is given to us for a certain gas and we need to find the specific heats ratio of that gas. Problem is as shown in the diagram below.


Solution

This process is adiabatic process and we need to write the given relation between pressure and temperature from the given format to the standard format so that we can solve and find the ratio of specific heats of the given gas as shown in the diagram below.


Problem

When a mono atomic gas expands at constant pressure, the percentage of the heat supplied that increases the temperature of the gas and in doing external work is how much is the problem and it is as shown in the diagram below.


Solution

We know that if volume is kept constant, work done is zero and corresponding specific heat is defined as per it and vice versa. When we write ratio of specific heats, it becomes the ratio of heat supplied to the change in the internal energy and internal energy change the temperature of the system. Thus we can find the percentage value basing on the specific heats ratio value as shown in the diagram below.



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Thermodynamics Problems with Solutions Three

We are solving series of problem in the topic thermodynamics. There are different types of process in thermodynamics. Isobaric process means pressure of the system is kept constant. Isothermal process means temperature of the system is kept constant. Adiabatic process means heat energy of the system is kept constant. In each case, work done is different and we need to measure as per the given process. Zeroth law of thermodynamics is regarding temperature concept. First law of thermodynamics is regarding conservation of energy and second law of thermodynamics is regarding direction of flow of heat.

Problem

Three samples of same gas has initially same volume. Volume of each one is doubled and the process is different in each case. It is adiabatic, isobaric and isothermal respectively . if all the final pressures are equal, we need to know the initial pressures of the three and the problem is as shown in the diagram below.


Solution

The relation between pressure and volume is different in each case. Isobaric process means pressure remains constant and in that case, initial pressure is equal to final pressure. In the case of isothermal process, as temperature is constant, we can apply boyle’s law and find the final pressure in terms of initial pressure. Adiabatic process get the ratio of specific heats into the picture and taking that into consideration, we can solve the problem as shown in the diagram below.


Problem

The pressure inside a tyre at one temperature is given to us as four atmospheric pressure and we need to find the temperature when the tyre bursts suddenly. Problem is as shown in the diagram below.


Solution

As the tyre bursts suddenly, it is going to be a adiabatic process and the heat energy of the system remains constant. We need to apply and find the relation between temperature and pressure using the pressure and volume relation of the adiabatic process and simplify the problem as shown in the diagram below.


Problem

There are two cylinders with the same ideal gas at the same temperature. There are pistons on both of them and their initial temperature is same. If one piston is allowed to move freely and the other is fixed, we need to compare the temperature of the second case when compared with the first case. Problem is as shown in the diagram below.


Solution

In the first case pressure is kept constant and in the second case it is the volume that is kept constant by sealing the system. Taking the respective specific heats and using the equation for the heat energy in each case, we can solve the problem as shown in the diagram below.


Problem

An ideal gas after going through four thermodynamic states has come back to its initial state. Heat energy in each case is given to us and work done in three cases is given to us as shown in the diagram below. We need to find the work done in the fourth case.


Solution

As the process is cyclic, there is no change in the internal energy of the system and as per the first law of thermodynamics, the heat energy supplied in this case will be the work done itself. By equating the total heat energy with proper sign to the total work done, we can solve the problem as shown in the diagram below.


Problem

PV diagram of an ideal gas is as shown and we need to measure the work done during a part as shown in the diagram below.


Solution

Work is said to be done when there is a change in the volume of the system other wise the work done is zero. In a PV diagram, work done is the area under the graph. We can use the shape and find the area of the graph to solve the problem.



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Thermodynamics Problems with Solutions One

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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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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