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Suppose we put a jar in front of a lamp (which mimics sunlight) and wait for the jar to heat up. When the temperature of the jar increases by few degrees, it stops increasing.

My thought is that as the jar absorbs heat, it also gives heat off to surroundings. When the heat absorbed equals the heat of surrounding area, the temperature of jar no longer increases.

However, I am not quite sure of this explanation. Is it correct?

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    $\begingroup$ Why does a car with a fixed amount of power accelerate to a particular speed then stop accelerating? $\endgroup$
    – Darren H
    Commented Apr 2, 2017 at 11:24

2 Answers 2

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It's all to do with the laws of thermal dynamics, in particular with how an object emits radiation with temperature: see the Stefan-Boltzmann law

You can see with metals, that once you heat them up they start to glow. Heat them up hotter, and they glow brighter - this shows that hotter things are brighter because as temperature increases further everything emits more radiation, so given a set power of a lamp, the jar you mention will eventually reach a thermal equilibrium, where it is taking heat equal to the amount it releases to the local environment.

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    $\begingroup$ In addition to radiating energy, the jar is releasing energy due to contact with the surrounding air, at a rate proportional to the difference between the temperature of the jar and the air. $\endgroup$
    – Carmeister
    Commented Apr 1, 2017 at 21:15
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You should distinguish different words:

  • "Heat" - a form of energy
  • "Power" - a rate of transfer of energy (an amount of heat per second)
  • "Temperature" - depends on the heat, mass, and heat capacity of the object

One of the "laws" is that "heat won't pass from a cooler (low-temperature) body to a hotter (high-temperature) body". If you put the two bodies together (without insulation) then heat will pass from the hotter to the cooler. When they reach the same temperature then the two bodies are in thermal equilibrium with each other.

The situation with the lamp isn't quite like that. You might have:

  • A radiant lamp
  • Radiating heat into a jar
  • Which is surrounded by cold air

It's not true that the jar and the air reach the same temperature; what is true is that thermal equilibrium is reached, when the rate at which heat is absorbed by the jar (from the lamp) equals the rate at which the jar loses heat (via radiation and conduction and convection) to the colder air which surrounds it.

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  • $\begingroup$ Heat is not a form of energy, a body has no "heat" but heating is a form of work $\endgroup$
    – hyportnex
    Commented Apr 1, 2017 at 20:57
  • $\begingroup$ I thought that heat was measured in Joules, and stored in a body (not unlike "potential energy" perhaps). Total heat in a body is calculated using the formula "temperature x mass x specific heat capacity". $\endgroup$
    – ChrisW
    Commented Apr 1, 2017 at 21:05
  • $\begingroup$ Work is also measured in joules. If "heat" was energy then a body that has energy would also contain "heat". Would all that heat equal the internal energy or just part of it? If you choose all then it is just another word for internal energy but then in the equation $dU=\delta Q + \delta W$ what is the difference between the two terms on the right? If "Q" is only a part of the internal energy then how can you tell what and how much part of it? $\endgroup$
    – hyportnex
    Commented Apr 1, 2017 at 21:16
  • $\begingroup$ Your definition of "total heat" as "temperature $\times$ mass $\times$ specific heat capacity" as a meaningful physical quantity can only make sense if the specific heat capacity is temperature independent. Yes you can write it as an integral $ChrisW(Q) = \int_0^T mc(T')dT'$ but going beyond that does it mean anything else rather than just being another word for "temperature" $T$ of the body? Because as written it has nothing to do with either the 1st or the 2nd law of thermodynamics. What will happen if $c(T)$ depends also on the external pressure or an external magnetic field? $\endgroup$
    – hyportnex
    Commented Apr 1, 2017 at 21:21
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    $\begingroup$ Anyway, for what it's worth, my answer remains (it's the jar's temperature, not the jar's internal "heat", that matters): in this case (where there's a lamp generating new heat) equilibrium isn't when the jar "equals the heat of" (i.e. have the same temperature as) the air which surrounds it ... rather, equilibrium is when the jar's power input (incoming radiant heat being absorbed) matches the jar's power output (outgoing heat), which (the existence of outgoing power) probably happens when the jar is hotter (higher temperature than) its surrounding air. $\endgroup$
    – ChrisW
    Commented Apr 1, 2017 at 21:44

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