There are two objects of the same material, similar geometry but different dimensions and same temperature. The larger object is supposed to contain more quantity of heat because it has larger volume.

Now I know that the quantity of emitted thermal radiation is directly proportional to temperature, but is it also regarding to internal energy?

How do I know which object emits more thermal radiaion? (without measurement)

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    $\begingroup$ For similar shaped objects the larger object will emit more radiation, but this is because it has a greater surface area, not because of it's volume or internal energy. Note that for objects with a very different shape it may be that the object with the smaller volume and/or heat capacity can have a larger surface area and so may emit more radiation $\endgroup$ – By Symmetry Aug 14 '19 at 10:30
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    $\begingroup$ Objects can't contain heat. Heat is energy that is transferred due to temperature differences. $\endgroup$ – BioPhysicist Aug 14 '19 at 10:37
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    $\begingroup$ Per Aaron Stevens objects don't contain heat. They have internal energy. Suggest you edit your question and change the terms "quantity of heat" and "heat" to "internal energy" $\endgroup$ – Bob D Aug 14 '19 at 11:22
  • $\begingroup$ Aaron and Bob, there are contexts where objects do contain heat. It is standard in chemical engineering to establish a base temperature, and use that temperature to calculate the enthalpy, or heat content, of a give mass of a substance such as steam. Typically, zero enthalpy occurs for an ideal gas at zero Kelvin, but the basis can be defined at other conditions as needed because the concept of enthalpy is usually used to calculate enthalpy differences. $\endgroup$ – David White Aug 14 '19 at 13:48
  • $\begingroup$ @DavidWhite It may be that in chemical engineering objects do "contain" heat, but in thermodynamics there are no contexts where objects contain heat. Objects have internal energy. In thermodynamics heat is energy transfer due solely to temperature difference. $\endgroup$ – Bob D Aug 14 '19 at 22:42

@Andreas Sundstrom is correct that the equation for radiation does not depend on the volume (mass) of the radiating object. But that is because the equation assumes the temperature of the emitting body is constant.

As the body radiates heat its internal energy is reduced thus reducing the temperature of the body and the rate of radiant heating. How quickly this occurs depends, of course, on the mass of the body and its specific heat and the variables in the radiant heat transfer equation. I should think the temperature of a sheet of paper would would drop quickly due to its low mass (volume).

Bottom line, to the extent that the temperature of the radiating body depends on its internal energy and the rate of radiant heat transfer depends on temperature, the rate of radiant heat transfer depends on the volume of the emitting object.

Hope this helps

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If you just mean the blackbody radiation, then that the radiated power both depends on the object's surface area, $A$, and its temperature to the forth power, $T^4$, according to $P_{\rm rad} = \epsilon\sigma A T^4$, where $\epsilon$ is the object's emissivity (material dependent) and $\sigma=5.67\times10^{-8}{\rm\,W\,m^{-2}\,K^{-4}}$ is the Stefan-Boltzmann constant. Hence it's the surface area which determines the radiation.

Since two objects can have the same volume but vastly different surface areas (think of a sheet of paper which has a large surface area but a very small volume), the thermal radiation is not dependent on the volume. However if you have two objects of the same shape, e.g. two cubes or two spheres, then the surface area scales with the volume as $A\propto V^{2/3}$, which would then also affect the thermal radiation, but only for two object of the same shape.

To answer your final question: If you have two objects at the same temperature and material, then the one with the larger surface area will radiate more energy.

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  • $\begingroup$ You assumed a convex surface. $\endgroup$ – my2cts Aug 14 '19 at 10:52
  • $\begingroup$ @Andréas Thank you $\endgroup$ – user65035 Aug 14 '19 at 12:28
  • $\begingroup$ @my2cts Yes indeed, I did. Thanks for pointing that out. $\endgroup$ – Andréas Sundström Aug 15 '19 at 6:35

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