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

In summary: If I am correct:Q = Cp(T2-T1) andP = QVp/timeCp = specific heat andp = densityP = powerV =VolumeT2 = final tempT1 = initial tempIf I am correct:Q = Cp(T2-T1) andP = QVp/timeCp = specific heat andp = densityP = powerV =VolumeT2 = final tempT1 = initial tempThe mass (Vρ) belongs in your first equation; Q depends on how much air is being heated. It does not belong in the
  • #1
esw6838
3
0
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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  • #2
Welcome to the forum. Please read this

https://www.physicsforums.com/showthread.php?t=94379

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?
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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
 

1. What is Newton's Law of Cooling?

Newton's Law of Cooling is a mathematical law that describes the rate at which an object cools down or loses heat in a surrounding environment. It states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the temperature of the surrounding environment.

2. What is the equation for Newton's Law of Cooling?

The equation for Newton's Law of Cooling is dT/dt = -k(T-Ts), where dT/dt is the rate of change of temperature, T is the temperature of the object at a certain time, Ts is the temperature of the surrounding environment, and k is a constant that depends on the properties of the object.

3. How is Newton's Law of Cooling used in real-life situations?

Newton's Law of Cooling is used in various real-life situations, such as predicting the cooling of hot beverages, determining the temperature of a body during a fever, and analyzing the cooling of molten lava or metal in industrial processes. It is also used in weather forecasting and climate studies to analyze the cooling of Earth's surface.

4. What factors affect the rate of cooling according to Newton's Law of Cooling?

The rate of cooling according to Newton's Law of Cooling is affected by several factors, including the temperature difference between the object and its surroundings, the surface area of the object, the type of material the object is made of, and the presence of any insulating layers or barriers.

5. Are there any limitations to Newton's Law of Cooling?

Yes, there are certain limitations to Newton's Law of Cooling. It assumes that the object is in a medium with constant temperature and that the temperature difference between the object and its surroundings is small. It also assumes that the object has a uniform temperature throughout its volume and that the rate of heat transfer is directly proportional to the temperature difference. These assumptions may not hold true in all situations, leading to deviations from the predicted rate of cooling.

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