Why do we need to equate the rate of energy of the second plate to zero?

  • Thread starter Thread starter LCSphysicist
  • Start date Start date
  • Tags Tags
    Plates Series
AI Thread Summary
The discussion centers on the need to equate the rate of energy of the second plate to zero in the context of thermal equilibrium. The calculation of energy transfer involves the radiation flux from a black body, which has a transmission coefficient of zero. It is emphasized that for thermal equilibrium, the outgoing energy must equal the incoming energy, otherwise the temperature of the second plate would change. The conversation concludes that the assumption of equilibrium is necessary to justify the equality to zero. This understanding clarifies the relationship between energy conservation and thermal dynamics in the scenario presented.
LCSphysicist
Messages
644
Reaction score
162
Homework Statement
Print below
Relevant Equations
\n
1605881791524.png

1605881805475.png

First i computed the rate of energy wrt time of the second plate:
$$dq_{2}/dt = A \sigma ((373)^4/2 + (273)^4/2 - T_{2}^4)$$
Equaliting it to zero we get the answer. But i am not sure why do we need to equality it to zero.
The q arrow on the figure suggest me that it is conservation of energy (radiation flux), but i don't get why.
The black body has transmission coefficient ##\tau = 0##, so i don't understand why the flux of radiation need to be equal. The body could pretty well absorb the energy coming from 1 and emite an energy to 3 totally different, couldn't?
 
Physics news on Phys.org
For thermal equilibrium, outgoing =ingoing. If that were not true, the temperature ##T_2## would change
 
  • Like
Likes LCSphysicist
hutchphd said:
For thermal equilibrium, outgoing =ingoing. If that were not true, the temperature ##T_2## would change
Yes i imagined it, but as the questions says nothing about equilibrium, i discarded the possibility. But since you pointed it too, i think that's enough to close this topic.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top