Tuesday, April 18, 2017

The Transfer of Thermal Energy

http://billonscience.blogspot.com/2017/04/the-transfer-of-thermal-energy.html
Thermal energy is a relatively easy concept to grasp. Objects in nature have a temperature, and that is (by definition) the average kinetic energy of the molecules of the object. The higher the temperature, the higher the average kinetic energy. Naturally, the larger the object, the more molecules, which means more kinetic energy.

So, thermal energy is a function of how many molecules are present as well as what the temperature of those molecules are. For any given substance, the total thermal energy can be easily found. 

The total thermal energy would be how much energy could be given off if the object's temperature dropped to absolute zero. This theoretical idea is not practicalInstead, thermal energy is always described as the amount of energy given off or taken in as a result of some change in temperature. 

The formula for finding thermal energy given a temperature change is actually very easy, EXCEPT it relies on the idea of change in temperature. It is common to use T for temperature and likewise common (though it is possibly something new to many introductory students) to use the Δ as a symbol for "change."

Thus, Δis the symbol for change in temperature.

So, if something starts off at 100 C and ends up at 80 C, what is ΔT?

ΔT = Ti - Tf
ΔT = 100 - 80
ΔT=20

Finding Δis more about getting used to the symbol than anything else; it is the difference between starting and ending (initial and final) temperatures.

Now to find heat, we can use the formula:

Q = mcΔT

Where Q is heat or thermal energy, m is mass, c is a number called specific heat that you either look up or calculate, and Δis the change in temperature.

EXAMPLES:

How much heat is absorbed by 200 grams of water that starts at 25C and ends up at 30C, given that the specific heat of water is 4.2 J/g°C?

Q = mcΔT
Q = 200•4.2•(30-25)
Q = 200•4.2•5
Q = 4200 J


What is the specific heat of a metal that has a mass of 50 grams and changes temperature from 100C to 30C and gives off 4200 J of thermal energy?

Q = mcΔT
4200 = 50•c•(100-30)
4200 = 50•70•c
4200 = 3500c
4200/3500 = c
1.2 = c


TRANSFER OF ENERGY

Thermal energy transfer is a fairly simple concept. When (two or more) things are in the same environment (which we will call being in the same system) the molecules of the things will bump into each other until all of them have the same kinetic energy.

What does that mean? Those things with higher energy will give off their energy to the things with lower energy. After a period of equalization, everything in the system will have the same average kinetic energy.

So, suppose a hot piece of metal is put into a cup of cool water… The energy from the metal will be transferred into the water. The metal will cool off. The water will warm up.

The amount of energy given off will be equal to the amount of energy absorbed. This is a major law of physics! Energy cannot be destroyed. It can change forms, or be transferred from one thing to another, but it cannot just go away.

Therefore, if hot metal is put into cool water, the quantity of the energy lost is equal to the quantity of the energy gained. Thus, Qlost = Qgained.

SO… if you know how much energy was gained, then you know how much energy was lost. 

And?

Suppose you put a sample of hot metal (for which you know the starting temperature) into water. If you know the mass of the water and you know the temperature change of the water, using the known quantity for water's specific heat (4.2), you can calculate how much energy was gained. Then, if you know the energy gained by the metal, the mass of the metal, and the change of temperature of the metal, you can then calculate the specific heat.


SUMMARY THOUGHTS

Q = mcΔT
In a system, once equilibrium is reached everything in the system will have the same kinetic energy (temperature).
Energy lost from objects in a system will equal energy gained by other objects in the system until all of the objects have equal temperature.

No comments:

Post a Comment