Heat Loss Through a Box with a hole in it

AI Thread Summary
The discussion centers on calculating heat loss from a styrofoam box with a lightbulb inside, focusing on the contributions of conduction through the walls and convection through a hole. It highlights the complexity of the problem, noting that both conduction and convection must be considered, along with potential radiation effects. Participants express confusion about the implications of air's lower thermal conductivity compared to wood, questioning how this affects heat retention in a house. The conversation emphasizes that while thermal conductivity is important, mass transfer through convection plays a significant role in heat loss, especially when windows are opened. Ultimately, a circuit diagram or nodal analysis is suggested as a potential method for solving the heat loss problem.
dchau503
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This is mainly just a theoretical question: say you have a rectangular box made of styrofoam (one of those ice cooler things) and it is levitating in air at room temperature conditions. It also has one hole at the top of the container so that heat can get out of it. Inside the box is a lightbulb that provides heat of 10 W. What equations can I use to find out how much heat is dissipated out the container?

I know how to use the equations for conduction to calculate the heat loss from the wall but I'm confused on whether to use just convection equations for the hole or if I should also combine convection equations to the conduction of the wall. Any help would be appreciated. I'm also thrown off by the fact that air has a lower thermal conductivity than wood, which doesn't make intuitive sense because that would make it seem like you would be more insulated in the outside than if you were in a house.
 
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dchau503 said:
This is mainly just a theoretical question: say you have a rectangular box made of styrofoam (one of those ice cooler things) and it is levitating in air at room temperature conditions. It also has one hole at the top of the container so that heat can get out of it. Inside the box is a lightbulb that provides heat of 10 W. What equations can I use to find out how much heat is dissipated out the container?

At steady state 10 Watts are dissipated.

How much is lost out the walls vs out the hole is a very difficult problem to solve.
There will be radiation to the walls and out the hole, filament temp. is high so it's probably significant. Conduction to the air in the box. Convection to the walls and out the hole. convection and radiation (probably insignificant) at the outer walls etc etc. I don't think this could be solved analytically, if it could it'd take an iterative solution.

[/QUOTE]I'm also thrown off by the fact that air has a lower thermal conductivity than wood, which doesn't make intuitive sense because that would make it seem like you would be more insulated in the outside than if you were in a house.[/QUOTE]
I can't parse the italics.
What's surprising about air having lower thermal conductivity than wood? Generally, solids are much better conductors of heat than gasses. that's why you can put your hand very close to a hot surface without getting burnt..
 
How about if we just make this problem simpler: i.e. we can ignore radiation, there's just one rectangular slab of styrofoam with a hole in it. If you're just armed with convection and conduction, how would you calculate the heat loss if 10 W is applied equally throughout one side of the slab?

Also, it's confusing that air has a lower thermal conductivity because say I live in a wooden house that is at 70 degrees and the outside weather is -20 Celsius. If wood has a higher thermal conductivity, then that would mean that the temperature of the house would drop down much more quickly than if I were to have a house made of air (so no house) at the same initial conditions. I'm sure I'm ignoring something pretty important in this case though and it's driving me nuts.
 
In the winter try opening all of the windows. That gets you closer to a house "made of air". Why does it get cold inside?

BoB
 
It gets cold inside because of how heat goes from hot to cold. Because the thermal conductivity of air is lower than that of wood, I assume that the heat loss to the outside occurs at a slower pace when I open the windows than when I keep the (wooden) windows shut though. What am I missing?
 
dchau503 said:
It gets cold inside because of how heat goes from hot to cold. Because the thermal conductivity of air is lower than that of wood, I assume that the heat loss to the outside occurs at a slower pace when I open the windows than when I keep the (wooden) windows shut though. What am I missing?

I take it you've never opened a window on a cold day?

The difference is that closed windows prevent mass transfer, that's why we have windows! to let light in without mass transfer. Thermal conductivity doesn't matter one bit if cold air is blowing straight in through open windows.
Even if the air outside is completely still there will be mass transfer via convection currents. That is, the warm air in your house will float up right out your windows, and cool air will be drawn in, no conduction necessary.
 
I think a circuit diagram (nodal analysis) is the way to go.
 
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