Thermal Energy

Work in progress by KwangJun Jung

Thermal energy is energy possessed by an object or system due to the movement of particles within the object or the system.

The Main Idea

All objects are made up of numerous particles or molecules. The constant, random motion of these atoms or molecules is what we associate thermal energy with. Thermal energy is a component of internal energy, but is unrelated to the vibrational and rotational energy of a solid's atoms.. When we say change of "thermal" energy, we mean that it is that part of the internal energy that is associated with a temperature change. Thermal energy is quantified using temperature. This quantification describes the approximate average thermal kinetic energy present in all of the atoms or molecules in the system. In many situations it isn't possible to say how much of the internal energy is thermal, but if the heat capacity is known, we can use a thermometer to measure a change in the thermal energy.

A Mathematical Model

Change in thermal energy can be calculated by using the following mathematical equation. [math]\displaystyle{ {Q = m*C*dT} }[/math] where [math]\displaystyle{ {Q} }[/math] is the thermal energy in Joules, [math]\displaystyle{ {m} }[/math] is the mass of an object in grams, [math]\displaystyle{ {C} }[/math] is the object's specific heat capacity, and [math]\displaystyle{ {dT} }[/math] is the change in object's temperature in Celsius.

The Kinetic Molecular Theory of Matter

The Idea of Thermal energy is derived from the Kinetic Molecular Theory of Matter. This theory describes why matter exists in different phases and provides a description of the interactions and properties of atoms through the use of ideas that are generally applied to larger systems. According to the Kinetic molecular Theory, all matter is comprised of lots of smaller molecules or atoms that are constantly moving. The type of motion of these particles is a result of the thermal energy present and determines whether the substance is in a gaseous, liquid, or solid state. When energy is introduced or lost from a material, the resulting change in motion of the individual particles can cause a phase change occur for the substance. There are small spaces between the atoms that make up matter and as the thermal energy of a substance increases, these spaces get progressively larger and begin to overcome the intermolecular forces present.

Examples

Simple

500g of water was heated from the initial temperature of 20°C to 50 °C. What is the change in thermal energy of water? (Heat capacity of water is 4.2J/g*°C)

1. List given things and equation

[math]\displaystyle{ Q = m*C*dT }[/math]

[math]\displaystyle{ Q=?, m=500g, C=4.2J/g*°C, dT={T}_{f}-{T}_{i}, {T}_{f}=50°C, {T}_{i}=20°C }[/math]

2. Plug the numbers into the equation and find the answer

[math]\displaystyle{ Q = 500g*4.2J/g*°C*(50°C-20°C) }[/math]

[math]\displaystyle{ Q = 63000J }[/math]

Middling

400g of water with initial temperature of 90°C (specific heat of 4.2 J/g*°C) are poured into an aluminum pan whose mass is 800g with initial temperature of 20°C (specific heat of 0.9 J/g*°C). After a short time, what is the temperature of the water?

1. Write the equation and list the knowns and unknowns

[math]\displaystyle{ {{dE}_{water}+{dE}_{pan} = 0}, dE = m*C*dT }[/math]

[math]\displaystyle{ {m}_{water} = 400g,{m}_{pan} = 800g, {C}_{water} = 4.2J/g*°C {C}_{Aluminum} = 0.9J/g*°C, {T}_{i water} = 90°C, {T}_{i pan} = 20°C, {T}_{f} = ? }[/math]

2. Plug the numbers into the equation

[math]\displaystyle{ 0 = 400g*4.2J/g*°C*({T}_{f}-90°C) + 800g*0.9J/g*°C*({T}_{f}-20°C) }[/math]

3. Solve for [math]\displaystyle{ {T}_{f} }[/math] and find its value

[math]\displaystyle{ {T}_{f}=69°C }[/math]

Difficult

500g of water with initial temperature of 87°C (specific heat of 4.2 J/g*°C) are poured into an aluminum pan whose mass is 800g with initial temperature of 22°C (specific heat of 0.9 J/g*°C). Then you place the pan on a hot electric stove. While the stove is heating the pan, you stir the water doing 26000J of work, rising temperature of a system to 82.5 °C. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan?

1. Write the equation and find the final temperature when water and pan first reached thermal equilibrium.

[math]\displaystyle{ {{dE}_{water}+{dE}_{pan} = W + Q}, dE = m*C*dT }[/math]

[math]\displaystyle{ W=0, Q=0, {m}_{water}=500g, {m}_{pan}=800g, {C}_{water}=4.2 J/g*°C, {C}_{aluminum}=0.9 J/g*°C, {T}_{1water}=87°C, {T}_{1pan}=22°C, {T}_{2}=? }[/math]

2. Plug the numbers into the equation and find the final temperature

[math]\displaystyle{ {dE}_{water}= 500*4.2*({T}_{2}-87) }[/math]

[math]\displaystyle{ {dE}_{pan}=800*0.9*({T}_{2}-22) }[/math]

[math]\displaystyle{ {dE}_{water}+{dE}_{pan}=0 }[/math]

3. Solve for [math]\displaystyle{ {T}_{f} }[/math] and find its value

[math]\displaystyle{ {T}_{f} = 70.404 }[/math]

4. Write the equation and list the knowns, repeating the steps above (final temperature above now becomes initial temperature)

[math]\displaystyle{ {{dE}_{water}+{dE}_{pan} = W + Q}, dE = m*C*dT }[/math]

[math]\displaystyle{ W=26000J, Q=?, {m}_{water}=500g, {m}_{pan}=800g, {C}_{water}=4.2 J/g*°C, {C}_{aluminum}=0.9 J/g*°C,{T}_{2}=70.404°C, {T}_{3}=82.5°C }[/math]

5. Plug the numbers into the equation

[math]\displaystyle{ 500*4.2*(82.5-70.404)+800*0.9*(82.5-70.404) = 26000 + Q }[/math]

6. Solve for Q and find its value

[math]\displaystyle{ Q = 8110J }[/math]

Connectedness

How is this topic connected to something that you are interested in?

I find thermal energy interesting because it relates into thermal dynamics which describes how thermal energy is converted to and from other forms of energy and how it affects matter. I just find it really interesting how simply differentiating a temperature of a system can bring various changes to that system.

How is it connected to your major?

I am a 1st year student in Mechanical Engineering. Under mechanical engineering there is a branch of physics called Thermodynamics which deals with heat and temperature and their relation to energy and work. Thermal Energy is one of the topic under Thermodynamics dealing with macroscopic variables especially internal energy.

Is there an interesting industrial application?

One of the interesting industrial application of thermal energy is industrial thermal energy storage. Thermal energy storage is a technology that stocks thermal energy by heating or cooling a storage medium so that the stored energy can be used at a later time for heating and cooling applications and power generation.

Thermal Equilibrium

Like all systems in nature, thermal energy works to reach equilibrium. When to substances of differing temperature come into contact, the substance of greater thermal energy will do microscopic work on the other substance until the two are at the same temperature. Atoms of the higher energy substance will collide will those of the lower energy substance at the boundary. Kinetic energy will be transferred and distribute throughout. Given enough time, this process of propagation of kinetic energy will continue until equilibrium is reached. The transfer of thermal energy is referred to as heat.

History

Thermal energy was discovered by a man named James Joule in the 1840s. Careful experiments show that the temperature increase of an object and its surroundings due to friction is directly related to the amount of mechanical energy lost. Joule carried one of the most famous experience demonstrating this fact. Joule hung some weights from pulleys so that as they fell, they turned a paddle-wheel apparatus immersed in a bucket of water. The friction between the paddles and the water raised the water’s temperature by an amount that was directly proportional to the distance that the weights fell. In our modern system of units, Joule found that raising the temperature of a kilogram of water by one degree Celsius required a loss in mechanical (gravitational) energy of approximately 4200 joules. Joule therefore proposed that this mechanical energy is not actually lost, but converted into a new type of energy: thermal energy, which manifests itself as an increase in temperature.

See also

The Energy Principle

Kinetic Energy

Work

Further reading

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 7

http://physics.weber.edu/schroeder/eee/chapter3.pdf

External links

http://www.eschooltoday.com/energy/kinds-of-energy/what-is-thermal-energy.html

http://study.com/academy/lesson/what-is-thermal-energy-definition-examples.html

References

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 7

http://physics.weber.edu/schroeder/eee/chapter3.pdf