Thermodynamics: Fan in a room

[teme92]

Homework Statement

A person living in a 4m x 5m x 5m room turns on a 100-W fan before he leaves a warm room at 100kPa, 30 degrees Celsius, hoping that when he returned in 5 hours, the room would be cooler. Disregarding any heat transfer determine the temperature of the room when he comes back. ( Assume the air to be an ideal gas with molar heat capacity C_v=3R/2).

Homework Equations

I couldn't get any of the latex code to work in the new website format.

R=k_B/N_A
pV=nRT
deltaU=Q+W

The Attempt at a Solution

So I calculated the energy put into the system by the fan and got: E=100(60)(60)= 1,800,000 J

The volume of the room doesn't change but I know the pressure will increase so therefore the temperature will increase. I'm having trouble finding out how to find this though. Would anyone be able to point me to the next step? Any help is much appreciated.

 

[Chestermiller]

Since there is no heat transfer into or out of the room, what is delta U? How is deltaU related to the temperature change?

Chet

 

[teme92]

deltaU is 0? Well Q=mCdeltaT, am I on the right track with that?

 

[Andrew Mason]

deltaU is 0? Well Q=mCdeltaT, am I on the right track with that?

Delta U = Q + W where W is the work done on the gas. Is any work being done on the gas? (it says to ignore Q). That will tell you what happens to U.

AM

 

[Chestermiller]

deltaU is 0? Well Q=mCdeltaT, am I on the right track with that?

No. delta U is not zero. The fan is doing work on the gas. How is delta U related to the work W when Q is zero?

 

[teme92]

So deltaU = W then?

Sorry for the late reply and thanks for all the help!

 

[mfb]

E=100(60)(60)= 1,800,000 J

Don't forget units and the 5 hours on the left side.

Pressure will be constant - you cannot pressurize a real room, air will escape through various openings.

 

[teme92]

Oh I forgot to multiply by 5. So E = 9,000,000 J

 

[mfb]

No you did not forget it (otherwise your answer would not fit), you just forgot to write it down.

 

[Chestermiller]

So deltaU = W then?

Sorry for the late reply and thanks for all the help!

Yes.