How Does Gas Pressure Affect Heat Conduction in a Vacuum Flask?

In summary, the vacuum flask of radius r and length l is made of two concentric cylinders separated by a narrow gap. The liquid in the flask is at 60 degrees Celcius and the air outside is at 20 degrees Celcius. The rate of heat loss by conduction is estimated to be κ = 1/3 Cmolecule n λ <v>.
  • #1
Fek
16
0

Homework Statement



A vacuum flask of radius r and length l consists of two concentric cylinders separated by a narrow gap containing a gas at pressure 10^-2 Nm^-2. The liquid in the flask is at 60 degrees Celcius and the air outside is at 20 degrees celcius. Estimate the rate of heat loss by conduction.

Homework Equations



κ = 1/3 Cmolecule n λ <v>

λ = ( sqrt(2) * n * σ) ^ -1

p = 1/3 nm <v2>

pV = nkbT

J = - κ ∇T

where n is molecules per unit volume.

The Attempt at a Solution



κ is independent of pressure so the equation is still valid.

Cv per mol is 3/2 R for an ideal gas. So if I divide by avagadro's I will get heat capacity per molecule.

<v> = sqrt( 8KBT / m pi)

The gas in the "vacuum" is air, so mainly nitrogen, so m = 14.

For the temperature of the "vacuum" gas, as it is definitely between 293K and 333K it does not matter exactly.

So now we just need the mean free path. We can estimate the collision cross section as 2a^2 where a is the atomic radius.

Then we can find n either by using the equation for pressure or the ideal gas equation
n = p / KBT
Still left with ∇T unknown which seems like a dead end.
 
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  • #2
You're trying to find J. Do you know how to determine ∇T if the gap is small and the temperature is assumed to vary linearly with radius? How does the mean free path compare with the gap between the cylinders?

Chet
 
  • #3
Thanks for your response Chet.

∇T ≈ ΔT / s , where s is the gap between cylinders if we assume that ∇T is constant. However s is not given.

The expression for κ assumes the mean free path is much less than the size of the container, so the molecules collide with each other much more frequently than the container - I now see that this is not the case we can't use that expression.

However without using it I am struggling to see a method.

I suppose you could assume that the particles simply bounce back and forth across the gap and gain/lose internal energy each time they come into contact with a wall and transfer it that way. However it would be difficult to estimate the energy gained in each contact and the route between walls taken by the particle.

I also know Newton's Law of cooling: AH = hA ΔT , but I have not learned about what determines h.
 
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  • #4
Fek said:
Thanks for your response Chet.

∇T ≈ ΔT / s , where s is the gap between cylinders if we assume that ∇T is constant. However s is not given.

The expression for κ assumes the mean free path is much less than the size of the container, so the molecules collide with each other much more frequently than the container - I now see that this is not the case we can't use that expression.

That's what I was afraid of. Sorry, wish I could help you, but I don't have any experience in this area.

Chet
 
  • #5


I would suggest using the Fourier's law of heat conduction to estimate the rate of heat loss in this scenario. This law states that the rate of heat transfer through a material is proportional to the temperature gradient and the cross-sectional area, and inversely proportional to the material's thermal conductivity. In this case, the material is the gas inside the vacuum flask and the temperature gradient exists between the liquid at 60 degrees Celsius and the air outside at 20 degrees Celsius.

To apply this law, we can first calculate the temperature difference between the inner and outer surfaces of the flask, which is 60 - 20 = 40 degrees Celsius. Then, we can use the formula for thermal conductivity (κ) you provided to estimate the value. As you mentioned, κ is independent of pressure, so we can use the ideal gas equation to calculate the number of molecules (n) per unit volume. Once we have these values, we can use the formula for heat transfer (Q) through a material to estimate the rate of heat loss by conduction, which is given by Q = -κA∇T, where A is the cross-sectional area of the flask and ∇T is the temperature gradient.

It is important to note that this is only an estimation and there may be other factors that could affect the rate of heat loss, such as convection or radiation. However, using these equations and assumptions can provide a reasonable estimate for the rate of heat loss by conduction in this scenario.
 

Related to How Does Gas Pressure Affect Heat Conduction in a Vacuum Flask?

1. What is heat loss by conduction?

Heat loss by conduction is the transfer of heat between two objects that are in direct contact with each other. It occurs when there is a temperature difference between the two objects and heat is transferred from the warmer object to the cooler one.

2. How does heat loss by conduction happen?

Heat loss by conduction happens when molecules in the warmer object vibrate and collide with the molecules in the cooler object, transferring heat energy. This process continues until both objects reach an equal temperature, known as thermal equilibrium.

3. What factors affect heat loss by conduction?

There are several factors that can affect heat loss by conduction, including the temperature difference between the objects, the type of material the objects are made of, the surface area of contact, and the distance between the objects.

4. How can heat loss by conduction be reduced?

Heat loss by conduction can be reduced by using materials with low thermal conductivity, such as insulators, which can slow down the transfer of heat between objects. Additionally, increasing the distance between the objects and reducing the surface area of contact can also help reduce heat loss by conduction.

5. What are some real-life examples of heat loss by conduction?

Some real-life examples of heat loss by conduction include holding an ice cube in your hand and feeling it melt, placing a hot pan on a kitchen counter and feeling the counter become warm, and walking barefoot on a cold tile floor and feeling your feet become cold.

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