Calculating Heat Dissipation in a 100-W Lightbulb with 3.2cm Glass Bulb

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A 100-W lightbulb produces 95 W of heat, which is dissipated through a glass bulb with a radius of 3.2 cm and a thickness of 0.60 mm. To calculate the temperature difference between the inner and outer surfaces of the glass, the thermal conductivity of glass, which is 0.84 J/(s*m*C), is essential. Understanding how to apply this value with the bulb's dimensions and thickness is crucial for determining the rate of heat flow through the glass. The discussion emphasizes the need for clarity on using thermal conductivity in heat dissipation calculations. Accurate calculations will help in understanding the thermal performance of the lightbulb.
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A 100-W lightbulb generates 95 W of heat, which is dissipated through a glass bulb that has a radius of 3.2cm and is 0.60mm thick.

What is the difference in temperature between the inner and outer surfaces of the glass?

i am completely lost and need help
 
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npr2036 said:
A 100-W lightbulb generates 95 W of heat, which is dissipated through a glass bulb that has a radius of 3.2cm and is 0.60mm thick.

What is the difference in temperature between the inner and outer surfaces of the glass?

i am completely lost and need help

You need to know the thermal conductivity of the glass.
 
the thermal conductivity of glass is 0.84 J/(s*m*C)

what do i do with this
 
npr2036 said:
the thermal conductivity of glass is 0.84 J/(s*m*C)

what do i do with this
Suppose you knew the temperature difference, the dimensions of the glass bulb and the glass thickness. How would you use that to find the rate of heat flow through the glass?
 

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