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Radiant heat transfer and specific heat

  • #1
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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.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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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.
 
  • #3
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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?
 
  • #4
TSny
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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
 
  • #5
36
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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.
 
  • #6
TSny
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Gold Member
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2,743
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.)
 
  • #7
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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?

$$ \frac{S_{Al}}{S_{Cu}}=\frac{t_{Al} \rho_{Cu}}{t_{Cu} \rho_{Al}} =2.32$$
 
  • #8
TSny
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Gold Member
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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.
 
  • #9
36
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Ok. Thank you very much!
 

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