Heat engines working over time

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Matt Armstrong

Homework Statement


An office with a volume of 52.5 m^3 uses a heat engine that outputs 1200 W. Suppose that, initially, the office has a pressure of 1 atm and 20 degrees Celsius before the heater is turned on. After 10 minutes (6000 seconds), what is the final air temperature?

Homework Equations


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1 atm = 101325 Pa

20 + 273 = 293 K

Q = mc*ΔT ?

The Attempt at a Solution



At first, I thought this was a simple heat problem. I used the density of air and the volume to get a mass of 64.3125 kg. However, our book does not list the specific heat capacity of air, and I feel the answer I get may be a little contrived if I used values found on Google instead of using the variables given in the problem. However, I feel I may be using the wrong equation. I searched through my textbook and lecture notes, but I couldn't find any equations that I thought were relevant to the problem or had variables I could solve for with the given information. What is the equation I am supposed to be using here?
 
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Chestermiller said:
You’re using the right equation. What is the molar heat capacity at constant volume of an ideal diatomic gas in terms of the ideal gas constant R?

I used 5/2*R for C_v. However, this does not line up with my professors predicted temperature rise. The rise I got was less than a degree Celsius, but the answer we are supposed to get is between 10 to 20 degrees Celsius (though he writes that it is based on 'terrible assumptions' due to how heat works with any environment)
 
Matt Armstrong said:
I used 5/2*R for C_v. However, this does not line up with my professors predicted temperature rise. The rise I got was less than a degree Celsius, but the answer we are supposed to get is between 10 to 20 degrees Celsius (though he writes that it is based on 'terrible assumptions' due to how heat works with any environment)
Let’s see your work.
 
Chestermiller said:
Let’s see your work.

My apologies for not including. With Q=mcΔT, where heat = 1200, mass = 64.3125 kg, c = 5/2 R and ΔT = (T_f - 20), my final answer was 20.898 rounded to the thousandths.