Thermodynamics - heat transfer

AT14.In summary, the change in radiated power from an object with a constant area and emissivity changing from 80 to 433 degrees Celsius can be found by taking the initial power and multiplying it by (2.239x10^11)/(εAT^14), where T is the initial temperature in Kelvin.
  • #1
urgent
17
0

Homework Statement


if the temperature of an object changes from 80 to 433 degrees Celsius, but the area and emissivity remains constant, by how much does radiated power increase?


Homework Equations


P=phi*A*e*T^4


The Attempt at a Solution


i tried to do this:
80^4-433^4. and get the temperature change...but i do not know where to go from here, please if someone could give good explanation about this, i got an exam in 2 days=/..thanks!=]
 
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  • #2
So the change is E=εA(T24-T14) (remember to change the temperatures to Kelvin).

Originally the power was E=εAT14

So what is the ratio of the change in power to the original power?
 
  • #3
so i did this:
706^4 - 353^4
=2.329*10^11
um now how do i find the ratio of the change from this?
 
  • #4
urgent said:
so i did this:
706^4 - 353^4
=2.329*10^11
um now how do i find the ratio of the change from this?

So you have ΔE= εA(2.239x1011)

Initially E = εAT14

So get E and just find ΔE/E
 
  • #5


As the temperature of an object increases, the amount of heat energy it radiates also increases. This is described by the Stefan-Boltzmann law, which states that the power radiated by an object is proportional to the fourth power of its absolute temperature (in Kelvin) and its surface area and emissivity (which is assumed to be constant in this problem).

Using the equation you provided, we can calculate the radiated power at the initial temperature of 80 degrees Celsius (353 Kelvin) and the final temperature of 433 degrees Celsius (706 Kelvin).

P1 = phi * A * e * T1^4 = phi * A * e * (353)^4
P2 = phi * A * e * T2^4 = phi * A * e * (706)^4

To find the increase in radiated power, we can subtract P1 from P2:

P2 - P1 = phi * A * e * (706)^4 - phi * A * e * (353)^4

= phi * A * e * (706^4 - 353^4)

= phi * A * e * (1.192 x 10^14 - 4.45 x 10^11)

= phi * A * e * 1.192 x 10^14

Therefore, the radiated power increases by approximately 1.192 x 10^14 times when the temperature increases from 80 to 433 degrees Celsius. This shows the significant impact that temperature has on heat transfer through radiation.
 

What is the definition of thermodynamics?

Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. It is concerned with the behavior of macroscopic systems and how they respond to changes in their surroundings.

What is heat transfer?

Heat transfer is the process by which thermal energy is exchanged between objects or systems. It can occur through three main mechanisms: conduction, convection, and radiation.

What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, but can only change forms. This means that the total energy in a closed system remains constant.

What is the difference between heat and temperature?

Heat is a form of energy that is transferred from one object to another due to a difference in temperature. Temperature, on the other hand, is a measure of the average kinetic energy of particles in a system.

What is the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. This means that in any process, there will always be some energy that is lost and cannot be recovered.

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