Calculating Heat Loss From a Room of 125m3 - Confusing Results?

AI Thread Summary
A room measuring 125 m3 at 30C experiences a temperature drop of 0.001C per second, leading to a calculated heat requirement of 144W to maintain that temperature. However, using a different equation for heat loss through the walls results in a loss of 300W, creating a discrepancy of 156W. The confusion may stem from inaccuracies in the constants used for specific heat capacity, density, or thermal transmittance. Additionally, the discussion highlights the potential oversight of radiation heat transfer, which could significantly impact overall heat calculations. Accurate measurements and considerations of all heat transfer modes are essential for resolving these discrepancies.
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A room is 5x5x5=125 m3 and holds a temp of 30C. On the other side of all the walls the air is 20C. I measure that the room temperature drops about 0.001C per second. If I now use this equation...

<br /> \dot Q = V \cdot C_P \cdot \rho \cdot \dot T<br />

...I get the amount of heat that needs to be added in order for the room to keep this temperature (I think), and it's about 144W. What I don't get is that if I now calculate how much heat goes through the walls using the following equation...

<br /> \dot Q = U \cdot A \cdot \Delta T<br />

...this comes to 300W, but that means that I'm actually losing 300-144=156 Watts more than the first equation tells me!

Am I getting something all wrong here?! This really confuses me, and all attempts to explain this to me will be appreciated.
 
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The two equations are valid. I assume you're doing this experiment on a room and finding some discrepencies, is that correct?

If so, I'd guess that one or more of the constants you're using is off. It would be very difficult to accurately describe the overall C_P, \rho or U accurately.

Another problem may be overall heat transfer. Are you neglecting radiation heat transfer? Seems to me there would be some significant contributions due to the sun's heating.
 
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