# Radiant heat transfer and specific heat

• BobaJ
In summary, the conversation discusses a problem involving the transfer of heat by radiation between a small body and the walls of a large cavity. The rate of heat transfer is given by a specific equation when the temperature difference between the body and the walls is small. The conversation also delves into calculating the time for the temperature of the body to change by a certain amount and the ratio of specific heats for two different materials. The equations used involve the Boltzmann constant, the density of the materials, and the specific heat of the objects. Some approximations are made and the emissivity of the objects is also taken into consideration.
BobaJ

## Homework Statement

I'm a little bit stuck with this exercise.A small body with temperature T and emissivity ε is placed in a large evacuated cavity with interior walls kept at temperature Tw. When Tw-T is small, show that the rate of heat transfer by radiation is

$$\frac{dQ}{dt}=4T_{W}^3A\epsilon \sigma (T_{W}-T)$$

If the body remains at constant pressure, show that the time for the temperature of the body to change from T1 to T2 is given by

$$t=\frac{C_{P}}{4T_{W}^3A\epsilon \sigma}\ln(\frac{T_{W}-T_{1}}{T_{W}-T_{2}}).$$

Two small blackened spheres of identical size, one of copper and the other of aluminum, are suspended by silk threads within a large hole in a block of melting ice. It is found that it takes 10 min for the temperature of the aluminum to drop from 276K to 274K, and 14.2 min for the copper to drop the same interval of temperature. What is the ratio of specific heats of aluminum and copper? (The densities of Al and Cu are 2.70 x 10^3 kg/m^3 and 8.96 x 10^3 kg/m^3 at 25°C, respectively.)

## Homework Equations

I think that all relevant equations at given in the problem.

## The Attempt at a Solution

I have already been able to solve the second part. So, I already proofed the equation to obtain the time.

For the first part, I start with: dQ/dt=Aεσ(TW4-T4). I thought for a moment to split (TW4-T4) so that it says (TW+T)(TW2+T2)(TW-T) but then I do not know how to get the desired result.

And in the third part I'm a little bit stuck with the calculations. I assume that it works out with the equation of the second part, transformed so that it says $C_P=$... But so that it is the specific heat I have to get it "per unit mass". And then I have to divide them to get the ratio of specific heats. So, the Boltzmann constant and $T_W^3$ would cancel out. But I don't know where the densities come into the game and how to go on. I asume that the densities have to do something with the area.

I would appreciate some help.

BobaJ said:
I thought for a moment to split (TW4-T4) so that it says (TW+T)(TW2+T2)(TW-T) but then I do not know how to get the desired result.
You only need an approximation that is accurate to first order in the small quantity ##(T - T_W)##. Since your last factor is this small quantity, you can use "zeroth order" approximation for the other three factors. Zeroth order means that you can neglect any difference between ##T## and ##T_W##.

Another approach is to work with the differential of the function ##f(T) = T^4##.

And in the third part I'm a little bit stuck with the calculations. I assume that it works out with the equation of the second part, transformed so that it says $C_P=$... But so that it is the specific heat I have to get it "per unit mass".
In the equation for the time ##t##, ##C_P## is the heat capacity of the object. You will need to relate the heat capacity of the object to the specific heat of the material.

BobaJ
So, I could multiplicate the first two expressions and substitute TW where there is T, because the difference is neglectable? Because then I would get the 4TW3 that I'm looking for.

To relate the heat capacity (C) with the specific heat (S) is it correct to use: S=C/m where m is the mass of the object?

BobaJ said:
So, I could multiplicate the first two expressions and substitute TW where there is T, because the difference is neglectable?
Yes

To relate the heat capacity (C) with the specific heat (S) is it correct to use: S=C/m where m is the mass of the object?
Yes

BobaJ
Ok, as I need a mass (m=ρ*V). I could relate the volume with the area (A) given in the equation. So that V=(r/3)A and that m=ρ(r/3)A. If I substitute this into the equation I would get $$S=\frac{12 T_{W}^{3}\epsilon \sigma t}{\rho r ln(\frac{T_{W}-T_{1}}{T_{W}-T_{2}})}$$.

After this I would have to calculate the ratio of the specific heat of the two materials. So the TW3 cancel out, also the Stefan-Boltzmann constant would disappear. Would the natural logarithmus also disappear because it is the same interval of temperature? The radius would also cancel, because it the spheres have the same dimensions.
What do I do about the emissivity ε? And the density is also at a temperature of 25°C although we are working at roughly 1°C.

BobaJ said:
Ok, as I need a mass (m=ρ*V). I could relate the volume with the area (A) given in the equation. So that V=(r/3)A and that m=ρ(r/3)A. If I substitute this into the equation I would get $$S=\frac{12 T_{W}^{3}\epsilon \sigma t}{\rho r ln(\frac{T_{W}-T_{1}}{T_{W}-T_{2}})}$$.
OK

After this I would have to calculate the ratio of the specific heat of the two materials. So the TW3 cancel out, also the Stefan-Boltzmann constant would disappear. Would the natural logarithmus also disappear because it is the same interval of temperature?
Yes.
The radius would also cancel, because it the spheres have the same dimensions.
Yes
What do I do about the emissivity ε?
Both spheres are said to be "blackened".
And the density is also at a temperature of 25°C although we are working at roughly 1°C.
You should be able to neglect the temperature dependence of the densities. (To be sure, consult a table of thermal coefficients of expansion for the two metals.)

BobaJ
So I take as emissivity the value 1 for both, because they are blackened and I don't have to worry about a specific value.

Would this be right?

$$\frac{S_{Al}}{S_{Cu}}=\frac{t_{Al} \rho_{Cu}}{t_{Cu} \rho_{Al}} =2.32$$

BobaJ said:
So I take as emissivity the value 1 for both, because they are blackened and I don't have to worry about a specific value.

Would this be right?

$$\frac{S_{Al}}{S_{Cu}}=\frac{t_{Al} \rho_{Cu}}{t_{Cu} \rho_{Al}} =2.32$$
Looks good. (I get a small difference in the third digit.) Compare to the values of specific heats that you can find in tables.

Ok. Thank you very much!

## 1. What is radiant heat transfer?

Radiant heat transfer is the transfer of heat energy through electromagnetic waves, such as infrared radiation. This type of heat transfer does not require a medium, unlike conduction and convection.

## 2. How does radiant heat transfer differ from conduction and convection?

Radiant heat transfer differs from conduction and convection in that it does not require a medium and can occur through a vacuum. Conduction and convection, on the other hand, require a medium such as a solid, liquid, or gas to transfer heat energy.

## 3. What is specific heat?

Specific heat is the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. It is a measure of the ability of a material to store thermal energy.

## 4. How does specific heat affect the transfer of heat?

Specific heat affects the transfer of heat by determining how much heat energy is required to change the temperature of a material. Materials with a higher specific heat will require more heat energy to increase their temperature, while materials with a lower specific heat will require less heat energy.

## 5. What factors can influence the rate of radiant heat transfer?

The rate of radiant heat transfer can be influenced by several factors, including the temperature difference between the objects, the emissivity of the objects, and the distance between the objects. The type of material and its surface properties can also affect the rate of radiant heat transfer.

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