Cooling trouble

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  • #1
Kelju Ivan
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Air conditioner cools the air in a room with a power of 8.4*10^6J/h. Volume of the room is 62.5m^3. At first the air is at the temperature of 30C and air pressure is 1atm, which stays a constant. How long does it take to cool the room to 25C? Note that the air shrinks while cooling down and since the pressure stays constant, more air (at 30C) flows in from the outside. Treat the air as a diatomic gas.

I am overwhelmed by this exercise. Normally I could make the equation (S is the integral sign) S Pdt = S nRdT (n=pV/RT), but now more air is flowing in, all the time warming the room and making the n change. How the hell does this work?
 

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  • #2
BLaH!
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Ok, so you know that both the temperature and number of particles change with time. The pressure and volume are both constants. So, we can right

[tex]p_{0}V_{0} = N(t)k_{b}T(t) \Rightarrow N(t) = \frac{p_{0}V_{0}}{k_{b}T(t)}[/tex]

Next, we can right the power as

[tex]P = \frac{dQ}{dt} = C_{V}\frac{dT}{dt}[/tex]

It says that you have a diatomic gas. We know that the heat capacity of a diatomic gas is

[tex]C_{V} = \frac{5}{2}Nk_{b}[/tex]

Substituting for the heat capacity and the number of particles,

[tex]P = \frac{5p_{0}V_{0}}{2k_{b}T(t)}\frac{dT}{dt}[/tex]

Thus,

[tex]P\Delta t = \frac{5p_{0}V_{0}}{2k_{b}} ln \frac{T_{final}}{T_{initial}}[/tex]
 
  • #3
Kelju Ivan
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That differs a bit from the answer we were shown, but I guess it does just the same thing, you just used Boltzman's constant instead of R.

In our version we had [tex] C_p=\frac{7}{2}R [/tex]. I'm not really sure when I should use 5/2 and when 7/2, since both are valid for [tex]C_p[/tex].

[tex]t=Q/P[/tex] where

[tex] Q=nC_p\Delta T [/tex] and [tex] n=\frac{pV}{RT}[/tex]
 

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