How Long Does It Take to Heat a Room Using Newton's Law of Cooling?

AI Thread Summary
To calculate how long it takes to heat a room using Newton's Law of Cooling, the specific heat and density of air are essential. The air-cooled condensing unit rejects 102,600 BTU/hr into a room with a volume of approximately 27,195 cubic feet. The calculations involve determining the heat energy required (Q) and the power output (P) to find the time needed to increase the temperature by 1 degree Fahrenheit. Using SI units, the time calculated for this temperature increase is approximately 50 seconds, although there are concerns about the volume conversion from cubic feet to cubic meters. Accurate conversions and calculations are crucial for precise results.
esw6838
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I have a problem where I have an air cooled condensing unit located in a large room within a building. The volume of the room is approximately 27,195 cubic feet. The air cooled condensing unit rejects 102,600 btu/hr to the space. Assuming the ambient air temperature of the space is 95 degrees fahrenheit, how can I calculate how long will it take to heat the volume of air 1 degree fahrenheit? Any help is appreciated. Thanks.
 
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What is the specific heat of air? What is the density of air? How is specific heat related to temperature change? What have you done to try to solve this problem?
 
If I am correct:

Q = Cp(T2-T1) and
P = QVp/time

Cp = specific heat and
p = density
P = power
V =Volume
T2 = final temp
T1 = initial temp
 
esw6838 said:
If I am correct:

Q = Cp(T2-T1) and
P = QVp/time

Cp = specific heat and
p = density
P = power
V =Volume
T2 = final temp
T1 = initial temp

The mass (Vρ) belongs in your first equation; Q depends on how much air is being heated. It does not belong in the second equation since it is already included in Q.
 
I first solve for Q. After I know Q, I can solve for time since I know the power is 102,600 btu per hour. I used SI units and came up with the following:

Cp = 1,006 J per KgK
p = 1.15 Kg per cubic meter
V = 481 cubic meters
T1 = 308.15K (95F)
T2 = 310.9K (100F)
P = 30,060W (102,600 btu per hour

I wind up with time = 50 sec. I thought it would take longer. Interesting.
 
esw6838 said:
I first solve for Q. After I know Q, I can solve for time since I know the power is 102,600 btu per hour. I used SI units and came up with the following:

Cp = 1,006 J per KgK
p = 1.15 Kg per cubic meter
V = 481 cubic meters
T1 = 308.15K (95F)
T2 = 310.9K (100F)
P = 30,060W (102,600 btu per hour

I wind up with time = 50 sec. I thought it would take longer. Interesting.

I did not check all your conversions, but the volume looks a bit small to me. 1 meter is less than 3⅓ feet, and (3⅓ feet)³ is around 37. 27,195 ft³ should be around 740 m³
 
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