5
$\begingroup$

Consider a body (not blackbody) at temperature $T$ kept at surrounding of temperature $T_s$. Due to the temperature difference, heat will be lost from the body in the form of radiations from its surface area which in turn will change its body temperature. Is there any connection between the specific heat capacity and the heat lost?

$\endgroup$
  • 2
    $\begingroup$ The question is slightly unclear. In my answer, I interpreted this as being about the rate of heat loss at a certain temperature. $\endgroup$ – Pieter Jun 23 '18 at 15:24
4
$\begingroup$

Yes. Specific heat is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree). So an object with a large specific heat will absorb or radiate more heat to lower the temperature difference. In your example, the lower temperature mass will absorb photons from the warmer surroundings. It will absorb a number of photons such that the temperature difference is zero. But it will take more photons for an object with a higher specific heat.

$\endgroup$
  • $\begingroup$ The specific heat is quite irrelevant, and its value per mole is quite similar for most materials anyway. It is the total thermal mass that influences the cooling rate (but that is not how OP phrased the question). A thin foil cools much faster than a massive block. $\endgroup$ – Pieter Jun 24 '18 at 1:18
3
$\begingroup$

I'd like to add few more details and examples for above @jmh's nice and simple answer.

If we consider the efficiency(P) of an object which is not black, when the radiation is happening ,

 Radiate energy = (Emissivity) * (Stefan-Boltzmann constant) * (Temperature)4 * (Area)

Consider two objects which have different specific heat capacities. But they are in same temperature in two rooms(rooms are at normal room temperature) but the both objects are exceed of the room temperature.(Also consider the two objects have the same surface area.)

So, the both objects will try to get their temperature into the room temperature.

Think about the first second,the both objects emit the energy. At the first second both objects will emit same energy by the above formula.

But after that will they at the same temperature? Actually No due the specific heat capacity is different. The specific capacity upper object will decrease its temperature than the other.

So, after that second there will not having a same emission after that. The temperatures are different. So the energy emit for a second will be different after that moment. So, the both objects will get different times for come to the normal room temperature.

So, there is a connection between the specific heat capacity and the above formula, when this procedure is happening. But when we considering the formula individually, or within a moment it will not affect with the Stephan-Boltzmann Law Formula.

$\endgroup$
  • $\begingroup$ What specific heat then? Per mass, per volume, per mole? Much more than that is the thickness that influence the cooling rate. $\endgroup$ – Pieter Jun 24 '18 at 1:30
2
$\begingroup$

No. The rate of heat loss by radiation is only dependent on area and on the emissivity of the surface.

The total heat capacity of the body then influences how fast its temperature will change. But heat is not temperature in physics, it is energy.

$\endgroup$
  • 1
    $\begingroup$ Your express is right. It is valued for certain temperature. But it will change after a moment. Due to the specific heat capacity difference after the first second the temperature of two different heat capacity objects will be change. So the heat lost will be change after that moment due to the temperature is different in two objects. $\endgroup$ – Osal Thuduwage Jun 23 '18 at 15:30
  • $\begingroup$ @OsalThuduwage The specific heat capacity is quite irrelevant, and its value per mole is quite similar according to Dulong-Petit. It is the total heat capacity that influences the cooling rate. Are you then saying that thickness influences the heat lost? $\endgroup$ – Pieter Jun 24 '18 at 1:21

protected by Qmechanic Jun 23 '18 at 20:53

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.