# Newtons Law of Cooling

#### esw6838

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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#### esw6838

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

#### OlderDan

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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.

#### esw6838

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.

#### OlderDan

Homework Helper
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

"Newtons Law of Cooling"

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