Thermal Flask, heat conduction

The value of ##k## is on the order of 10-5 W/(m##\cdot##K) and the time taken is also on the order of 10^8 or 10^9 seconds. This seems more reasonable. In summary, the expression for thermal conductivity using kinetic theory is k = (1/3)C_vn<c>, where C_v is the heat capacity per molecule, n is the number of gas molecules per unit volume, and <c> is the average speed of the molecules. Using this formula and the given dimensions of the vacuum flask, the estimated heat loss per unit time is 7.14*10^-4 J/s. It would take approximately 3.7*10^
  • #1
unscientific
1,734
13

Homework Statement



Find an expression for thermal conductivity using kinetic theory.

Given a vacuum flask of these dimensions, find the heat loss per unit time.

Estimate the time taken for the water to cool down to 40 deg.

hvcqi8.png



Homework Equations





The Attempt at a Solution



Part (a)
[tex] k = \frac {1}{3} C_{mol} n λ <c> [/tex]

Part (b)

Estimate gap to be 0.5cm.

[tex] n = \frac {PV}{RT} =\frac {(10^{-2})(3.3*10^{-4})}{(8.31)(293)} = 1.4 * 10^{-9} [/tex]

Since the the molecular density is so low, we can effectively say the mean free path is the gap. [tex] λ ≈ 0.5 cm [/tex] (Not sure if this is right, but if the molecule would crash into the other end of the fask before it hits the molecule, then the mean free path should definitely be in the order of magnitude of the gap, right?)

Heat Capacity of air should be same order of magnitude as water;
[tex] C_{mol} ≈ 10^3 J K^{-1} [/tex]

[tex] <c> ≈ 400 m s^{-1} [/tex]

Estimate Temperature gradient ≈ ΔT/5cm) = 8000 K/m

Together, [tex] k ≈ 9.3*10^{-7} [/tex] So, [tex] P = JA = 4.69 * 10^{-4} [/tex]


Part (c)

Time taken = [tex] \frac {Q}{Power} = 1.79*10^8 s = 6 years [/tex]
 
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  • #2
The time taken is ridiculous, which is why I'm not sure if that's right.

Also, heat is conducted away by radiation.
 
  • #3
unscientific said:
[tex] n = \frac {PV}{RT} =\frac {(10^{-2})(3.3*10^{-4})}{(8.31)(293)} = 1.4 * 10^{-9} [/tex]

According to the formula for ##k## in the statement of the problem, ##n## is the number of gas particles per unit volume. Looks like you're letting ##n## be the number of moles of gas in the flask.

Since the the molecular density is so low, we can effectively say the mean free path is the gap. [tex] λ ≈ 0.5 cm [/tex] (Not sure if this is right, but if the molecule would crash into the other end of the fask before it hits the molecule, then the mean free path should definitely be in the order of magnitude of the gap, right?)

I'm not sure about what to do when λ is much greater than the gap. But your assumption seems to be reasonable. If you do assume that λ can be taken to be the thickness of the gap, then I think λ will cancel out by the time you calculate J.

Heat Capacity of air should be same order of magnitude as water;
[tex] C_{mol} ≈ 10^3 J K^{-1} [/tex]

In the formula for ##k##, ##C_{mol}## is the heat capacity "of a molecule". Is that what your ##C_{mol}## represents? There is a well-known formula for the molar heat capacity at constant volume of an ideal diatomic gas. From that you can get the heat capacity per molecule.
 
  • #4
TSny said:
According to the formula for ##k## in the statement of the problem, ##n## is the number of gas particles per unit volume. Looks like you're letting ##n## be the number of moles of gas in the flask.
I'm not sure about what to do when λ is much greater than the gap. But your assumption seems to be reasonable. If you do assume that λ can be taken to be the thickness of the gap, then I think λ will cancel out by the time you calculate J.
In the formula for ##k##, ##C_{mol}## is the heat capacity "of a molecule". Is that what your ##C_{mol}## represents? There is a well-known formula for the molar heat capacity at constant volume of an ideal diatomic gas. From that you can get the heat capacity per molecule.

the correct value of n should be [tex]n = 2.47*10^{18} [/tex]

From the ideal gas law: at constant volume, For N molecules of gas, [tex] dQ = dU = N C_{mol} dT = \frac {5}{2} Nk dT [/tex] So, [tex] C_v = \frac {5}{2}k = 3.45*10^{-23} [/tex] Using these data along with [tex] <c> = 400 m/s [/tex], we obtain:
[tex] κ = 5.68*10^{-5} [/tex]
[tex] Heat loss rate =7.14*10^{-4} J /s [/tex]
[tex] time taken = \frac {energy}{heat loss rate} = \frac {mCΔT}{7.14*10^{-4}} = 3.7*10^{8} s [/tex]

The time taken is astronomical!
 
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  • #5
unscientific said:
the correct value of n should be = [tex] \frac {Vol.}{(1.4*10^{-9})(6*10^{23}} [/tex]
The number density ##n## is the number of molecules per unit volume: ##n= N/V## where ##N## is the number of molecules and ##V## is the volume. From the ideal gas law ##PV = Nk_BT## you should be able to calculate ##n## from the pressure, temperature, and Boltzmann's constant ##k_B##.

From the ideal gas law: at constant volume, For N molecules of gas, [tex] dQ = dU = N C_{mol} dT = \frac {5}{2} Nk dT [/tex] So, [tex] C_v = \frac {5}{2}k [/tex] But this is per mol,

so per molecule = [tex] \frac {C_v}{6*10^{23}} = 3.4625*10^{-23}[/tex]

The molar heat capacity at constant volume for an ideal diatomic gas is ##\frac{5}{2} R##. But that leads to the same result as you have for the heat capacity per molecule.

Using these data along with [tex] <c> = 400 m/s [/tex], we obtain:
[tex] κ = 9.17*10^{-42} [/tex]

I think ##\kappa## should be on the order of 10-4 or 10-5 W/(m##\cdot##K)
 
  • #6
TSny said:
The number density ##n## is the number of molecules per unit volume: ##n= N/V## where ##N## is the number of molecules and ##V## is the volume. From the ideal gas law ##PV = Nk_BT## you should be able to calculate ##n## from the pressure, temperature, and Boltzmann's constant ##k_B##.
The molar heat capacity at constant volume for an ideal diatomic gas is ##\frac{5}{2} R##. But that leads to the same result as you have for the heat capacity per molecule.
I think ##\kappa## should be on the order of 10-4 or 10-5 W/(m##\cdot##K)

I realized I calculated the wrong value of n. I have adjusted my answers above, and I get the following:

[tex] k = 5.68*10^{-5}, time = 3.7*10^{8} s [/tex]
 
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  • #7
I get a similar result.
 

1. What is a thermal flask?

A thermal flask is a type of insulated container that is designed to keep the contents at a constant temperature, whether hot or cold. It is commonly used to keep beverages like coffee, tea, or soup hot, or to keep drinks like water or juice cold.

2. How does a thermal flask work?

A thermal flask works by utilizing a vacuum layer between two walls of the flask. This vacuum layer acts as an insulator, preventing heat transfer between the contents of the flask and the outside environment. Additionally, the inner wall of the flask is usually made of reflective material to further reduce heat transfer through radiation.

3. What is heat conduction?

Heat conduction is the transfer of thermal energy between two objects or substances that are in direct contact with each other. In the case of a thermal flask, heat conduction occurs between the contents of the flask and the inner wall of the flask, which is made of a material that is a poor conductor of heat.

4. Why is heat conduction important in a thermal flask?

Heat conduction is important in a thermal flask because it allows for the contents of the flask to be kept at a constant temperature. Without proper insulation and prevention of heat transfer, the contents of the flask would quickly reach the temperature of the surrounding environment and lose their desired temperature.

5. How do I choose the best thermal flask for my needs?

When choosing a thermal flask, there are a few factors to consider. These include the size and capacity of the flask, the material it is made of (stainless steel is a popular choice for its durability and insulation properties), and any additional features such as a built-in strainer or leak-proof lid. It is also important to read reviews and compare brands to find the best thermal flask for your specific needs.

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