Calculate Heat Loss from Copper Sphere of 8cm Diameter

In summary, a blackened copper sphere with a diameter of 8cm was suspended from the ceiling of a room kept at 25 degrees C. The sphere was then heated to a temperature of 147 degrees C. To calculate the net rate of heat loss by the sphere, it can be treated as a blackbody with an emissivity of 1 and the power radiated per unit area can be calculated using the formula \sigma T^4. The temperature in this formula refers to the temperature of the object itself, not the difference in temperature.
  • #1
EIRE2003
108
0
A copper sphere of 8cm diameter is blackened and suspended from the ceiling by a fine thread in a room kept at 25 degrees C. The sphere is then heated to a temp of 147 degrees C.
Calculate the net rate of heat loss by the sphere.

Im not given any constants such as stefans or the emissivity of the body, so how can this be calculated?
 
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  • #2
The sphere is copper (i.e. great thermal conductor) so you can assume its temperature is constant throughout its volume. Treat it as a blackbody (emissivity = 1) which allows you to calculate the power radiated per unit area as [itex]\sigma T^4[/itex].
 
  • #3
Thank You I'll try it out when i get up out of the scratcher!
 
  • #4
Is the Temperature the difference in the temperature or the temperature of the black body itself?
 
  • #5
It's the temperature of the object itself!
 

1. What is the formula for calculating heat loss from a copper sphere?

The formula for calculating heat loss from a copper sphere is Q = (4πkR2ΔT)/d, where Q is the heat loss in watts, k is the thermal conductivity of copper, R is the radius of the sphere, ΔT is the temperature difference between the sphere and its surroundings, and d is the thickness of the sphere.

2. How do you find the thermal conductivity of copper?

The thermal conductivity of copper can be found in reference tables or handbooks. It is typically around 400 watts/meter-kelvin at room temperature.

3. Is the diameter of the sphere or the thickness of the sphere more important in calculating heat loss?

The thickness of the sphere is more important in calculating heat loss, as it is directly included in the formula. The diameter of the sphere is used to calculate the surface area, which also affects heat loss, but it is not as significant as the thickness.

4. Can this formula be applied to other materials besides copper?

Yes, this formula can be applied to other materials, but the thermal conductivity value will be different. The formula can be modified to fit the specific material being used.

5. How accurate is this formula in calculating heat loss from a copper sphere?

This formula is accurate in calculating heat loss from a copper sphere as long as the assumptions are met, such as uniform thickness and temperature throughout the sphere. However, it is always recommended to verify the results with experimental data for more accurate results.

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