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Temperature of gas between window panes

  1. Dec 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A double paned window consists of two glasses and a gap between them. The inner glass is d1 = 6mm thick, the gap is D = 16mm thick and the outer glass is d2=4mm thick. The inside temperature (by the inner glass) is T1=23°C and the outside temperature is T2=-10°C. What is the temperature of the gas in the gap? Glass conductivity is k=0.96 W/(m*K)


    2. Relevant equations
    [itex]\frac{\Delta Q}{\Delta t} = kA\frac{\Delta T}{L}[/itex]


    3. The attempt at a solution
    I tried doing something like this: since we don't know area or time, but they are the same to both of the glasses, expressing them from the euation and then equating might work. But that's where I'm stuck: [itex]\frac{\Delta Q_1 d_1}{\Delta T_1 k} = \frac{\Delta Q_2 d_2}{\Delta T_2 k}[/itex]. Not sure what to do next .. Also, there might be a trick that we don't know what is the gas between the glasses - it's not necessarily air..
     
    Last edited: Dec 18, 2013
  2. jcsd
  3. Dec 18, 2013 #2
    Let q be the total rate of heat flow. For the first glass, [tex]ΔT_1=\frac{qd_1}{k_1A}[/tex]. What is the equation for the temperature change across the second glass layer? Assuming that the resistance to heat transfer of the gas is negligible (a highly questionable assumption), how are ΔT1 and ΔT2 related to the total temperature change (23 - (-10)) = 43? Use this to solve for q/A.
     
  4. Dec 18, 2013 #3
    Is the rate of heat flow the same for both of the glasses? If so, I get the answer (T-temperature of the gas) [itex]\Delta T_1 = T_1 - T; \Delta T_2 = T-T_2 => T=\frac{T_1 d_2 + T_1 d_1}{d_2 + d_1}[/itex].
    And if we don't assume that the heat transfer is negligible, I guess there's no solution then?
     
  5. Dec 18, 2013 #4
    The rate of heat flow is the same for both the glasses. Whatever heat flow goes through one glass has nowhere else to go but through the other. It's like an electric current flowing through two resistors in series. The temperature is like the voltage.

    Please show your work. I'd like to see the details of what you did.

    If you don't assume that the resistance to heat transfer through the gas is negligible, then you need to know the thermal conductivity of the gas to solve the problem. In that case, you treat the gas as just another layer in the stack, and there will be 3 ΔT's that you add up.

    Chet
     
  6. Dec 18, 2013 #5
    That's what I did:
    [itex]\Delta T = \frac{qd}{kA} =>\frac{q}{A}=k\frac{\Delta T}{d}[/itex]. Since both q and A are the same for both glasses, [itex]\frac{q}{A} = k\frac{\Delta T_1}{d_1} = k\frac{\Delta T_2}{d_2}[/itex]. From what I've written before (what equals delta T1 and T2, we get [itex]\frac{T_1 - T}{d_1} = \frac{T- T_2}{d_2} => T=\frac{T_1 d_2 + T_1 d_1}{d_2 + d_1}[/itex]
     
  7. Dec 18, 2013 #6
    Check the subscripts in you last equation. One of those T's should be a T2.
     
  8. Dec 18, 2013 #7
    Yes, I'm sorry - [itex]T=\frac{T_1 d_2 + T_2 d_1}{d_1 + d_2}[/itex]. Is this equation right?
     
  9. Dec 18, 2013 #8
    Yes.
     
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