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

[npr2036]

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

 

[haruspex]

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.

 

[npr2036]

the thermal conductivity of glass is 0.84 J/(s*m*C)

what do i do with this

 

[haruspex]

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?

 

[Nickie]

I would first start by calculating the surface area of the glass bulb using its radius and thickness. This will give us the total area through which the heat can be dissipated.

Next, I would use the formula Q=UAΔT, where Q is the heat dissipation rate, U is the heat transfer coefficient, A is the surface area, and ΔT is the temperature difference between the inner and outer surfaces. We already know Q (95 W) and A (calculated in the previous step) and we can assume a typical heat transfer coefficient for glass (U=5 W/m^2K).

Rearranging the formula to solve for ΔT, we get ΔT=Q/(UA). Plugging in the values, we get ΔT=95/(5*0.00192)= 98750 K.

This means that the temperature difference between the inner and outer surfaces of the glass bulb is 98.75 °C. This temperature difference is necessary for the heat to be dissipated from the bulb to the surrounding environment.