If every body emits radiation at a given frequency and temperature exactly as well as it absorbs the same radiation, how do objects heat up?
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3 Answers
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The premise is true if the object is in thermal equilibrium. See, for example, this Wikipedia article.
Besides radiation, heat can be transferred by conduction and convection.
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$\begingroup$ This is a good answer. A small elaboration: note that the premise is true only if the object is in thermal equilibrium (with the surrounding radiation field) and convection and conduction can be ignored. That condition is rarely if ever met. $\endgroup$– garypCommented Sep 4, 2014 at 21:19
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$\begingroup$ And before thermal equilibrium is met, there is more incoming radiation than outgoing, causing it to heat up. $\endgroup$– BMSCommented Sep 4, 2014 at 22:18
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First, a comment. Radiative heat transfer is oftentimes a non-factor for everyday objects encountered here on Earth. Radiative transfer is important for objects that can't exchange heat conductively or convectively, and for objects whose temperatures differ by a marked amount. That said, the rest of this answer will focus on radiative heat transfer.
If every body emits radiation at a given frequency and temperature exactly as well as it absorbs the same radiation, how do objects heat up?
Except for objects already in thermal equilibrium, that simply is not the case.
Consider two parallel flat plates in space. Convection and conduction can't happen because of the vacuum of space. We'll cover the backsides and edges of the plates with perfect mirrors so the only energy transfer is between the objects, and we'll cover the facing sides with a perfectly black material so the objects act like black bodies on those facing sides. Per the Stefan-Boltzmann law Plate A radiates energy to plate B at a rate $P_{A\to B} =A\sigma T_A^4$ while plate B radiates energy to plate A at a rate $P_{B\to A} =A\sigma T_B^4$. The net energy transfer to plate A is $\frac{dE_A}{dt} = A\sigma(T_B^4 - T_A^4)$ while the net energy transfer to plate B is $\frac{dE_B}{dt} = A\sigma(T_A^4 - T_B^4) = -\frac{dE_A}{dt}$.
The energy transfer is zero only if $T_A = T_B$. Otherwise, the outgoing energy from the warmer body exceeds the incoming energy from the cooler body (and vice versa). The warmer transfers heat to the cooler body. The process stops when the two bodies come into thermal equilibrium.
answered Sep 4, 2014 at 22:51
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A cold object heats up because at any given frequency it emits less energy and receives more energy than a hotter object.
In other words, at any given frequency an object is just as efficient an emitter as it is an absorber (with a black body being the most efficient), but hot objects emit more radiation than cold objects do at each frequency, so the energy transfer always goes from hot to cold.
