Newtons Law of Cooling applied to hot Neon and Hydrogen gas

[morrobay]

Homework Statement

1 mole of Hydrogen gas at 300 Kevin is contained in a thin walled copper container.
In container #2 there is 1 mole of Neon gas also at 300 Kevin.
The volume= 22.4 liters and surface area ,A = .476 m^2
The surroundings are at 100 deg Kelvin.
The specific heat of H2= 5 cal/mole (K) Specific heat for Neon = 3cal/mole (K)
The thermal conductivity (k) H2 = .0433 cal/sec For Neon = .0116 cal/sec

How many seconds for each gas to cool to 120 Kelvin ?

Homework Equations

Since the heat transfer equation , Q/t = -kA (T2-T1)/L
is with T2 at a fixed temperatue and in the above problem T2 , the temp of the gases,
are changing. I am applying k to the solution to Newtons law of cooling .
T(t) = T surr. + (Tgas - Tsurr)e^-kt

120K (t) = 100K + 200K e^-kt

The Attempt at a Solution

So 20K/200K= e^-kt
2.3 = kt
For Hydrogen t = 53 seconds ,
For Neon t = 198 seconds

note I originally put this problem in the physics section but it should qualify for chemical
thermodynamics. Hopefully the chemists here can be of more help !

 

[Mapes]

Check your units: if $kt$ is dimensionless (which it must be to take the exponential), $k$ must have units of s^-1. And why do your thermal conductivity values have units of power?

 

[morrobay]

Check your units: if $kt$ is dimensionless (which it must be to take the exponential), $k$ must have units of s^-1. And why do your thermal conductivity values have units of power?

The units are in power since I converted Watts to calories.
This k value is for: Q/t = -kA (T2-T1)/L
So there is a unit problem in appling it in the Newtons Law of Cooling solution.
The question now is are there tables of k values that can be used with Newtons solution,
or can it only be obtained by taking temperatures at t0 ,t1,t2 and solving for k. ?

Or since the thermal conductivity (k) above in Q/t = -kA (T2-T1)/L in units
(cal/sec)/meter^2)C deg/meter
And (k) in units 1/sec here: T(t) = Tsurr + (T initial - Tsurr) e^-kt
Are both related to heat transfer for a specific material can there be a conversion ?

 

[Mapes]

But thermal conductivity should have units of power length^-1 degree^-1...

 

[morrobay]

But thermal conductivity should have units of power length^-1 degree^-1...

Agree : Watts/(Kelvin ) (meter) multiplied by (delta T) ( Area m^2) divided by Length (m)
The units in post #3 were copied from a text, it is a confusing copy.
So with 1 watt = .238 cal/sec.
Thermal conductivity should be: cal/sec / kelvin * meter ( T2-T1) ( m^2) / m
Thats as in : Q/t = -kA (T2-T1)/L
With k itself in power/ degree * length
In Newtons law of Cooling: dT/dt = -k (T initial- T surr.)
With this k in 1/sec
And both of these constants are related to heat transfer in a specific material . Is There a conversion factor from the thermal conductivity k
to the law of cooling k ?

 

[Mapes]

Is There a conversion factor from the thermal conductivity k
to the law of cooling k ?

Not without more information, such as the thickness of the copper, or the convection coefficient on both sides. I don't see how the problem is solvable from the information given in post #1.

 

[nasu]

You cannot really solve this kind of problem just with Newton's law of cooling.
Even for a solid body, with no convection, it is a little more difficult than that.
The rate of heat transfer to the medium depends on the temperature difference between the surface of the material (your gas) and the temperature of the medium. But the evolution of the surface temperature depends on how fast can heat be transported from the inner volume to the surface.
You can imagine some extreme cases to see the point.
If the thermal conductivity is very high, then the temperature in the volume of the object is almost uniform and the whole material cools down as a whole. In this case you can find the solution just by using Newton's law of cooling. It probably works reasonably well for metals.

The other extreme is a very good (ideal) insulator. In this case the outer layer cools down to outside temperature but there is very little heat being transported from the inside to compensate for this. The inner volume stays at the same (initial temperature).

In real cases you are somewhere in between. There is a temperature gradient through the material. The question "how long it takes to cool down" is not even well defined.