Cooling rates for Hydrogen and Neon gas

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SUMMARY

This discussion focuses on the cooling rates of Hydrogen and Neon gases in a copper container at 300 Kelvin. The specific heat capacities are 5 cal/mole (K) for Hydrogen and 3 cal/mole (K) for Neon, with thermal conductivities of 0.0433 cal/sec and 0.0116 cal/sec, respectively. Using Newton's law of cooling, the time for Hydrogen to cool to 120 Kelvin is calculated as 265 seconds, while Neon takes 605 seconds. Alternative calculations using thermal conductivity in cal/sec yield 53 seconds for Hydrogen and 198 seconds for Neon.

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Homework Statement


1 mole of Hydrogen gas at 300 Kelvin is in a thin walled copper container vol 22.4 liters
surface area A = .476 M^2
In vessel #2 , 1 mole of Neon gas also 300 Kevin
The specific heat of H2= 5 cal/mole (K) Thermal conductivity (k) H2 = .0433 cal/sec. converts to .00866 deg K /sec
Specific heat Neon = 3 cal/mole (K) Thermal conductivity (k) Neon = .0116 cal/sec
converts to .0038 deg K/sec.
The surroundings are 100deg Kevin
How mant seconds for the gases to cool to 120 deg kelvin


Homework Equations


Since k is from Q/t = -kA (T2-T1)/L where T2 is fixed I am applying k to Newtons law of
cooling: T(t) = Tsurr + (Tvessal - T surr ) e^-kt



The Attempt at a Solution


For H2: 120K (t) = 100K + 200K e^-.00866t
2.3 = .00866 t t= 265 sec with kA t= 565 sec

for Neon 2.3=.0038t t= 605 sec with kA t =1292 sec
 
Last edited:
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morrobay said:

Homework Statement


1 mole of Hydrogen gas at 300 Kelvin is in a thin walled copper container vol 22.4 liters
surface area A = .476 M^2
In vessel #2 , 1 mole of Neon gas also 300 Kevin
The specific heat of H2= 5 cal/mole (K) Thermal conductivity (k) H2 = .0433 cal/sec. converts to .00866 deg K /sec
Specific heat Neon = 3 cal/mole (K) Thermal conductivity (k) Neon = .0116 cal/sec
converts to .0038 deg K/sec.
The surroundings are 100deg Kevin
How mant seconds for the gases to cool to 120 deg kelvin


Homework Equations


Since k is from Q/t = -kA (T2-T1)/L where T2 is fixed I am applying k to Newtons law of
cooling: T(t) = Tsurr + (Tvessal - T surr ) e^-kt



The Attempt at a Solution


For H2: 120K (t) = 100K + 200K e^-.00866t
2.3 = .00866 t t= 265 sec with kA t= 565 sec

for Neon 2.3=.0038t t= 605 sec with kA t =1292 sec

If converting k , the thermal conductivity, to Kevin/sec by way of dividing by the specific
heat was incorrect then this is a rework with k in cal/sec from table:
For Hydrogen, (t) 2.3 = .0433t t= 53sec
For Neon , (t) 2.3 = .0116t t=198sec

If this is not correct I would like to see someone work it out from the original problem
 

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