What is the thermal conductivity needed to maintain -80 degrees C for 24 hours?

Question

I am trying to understand thermal conductivity as it realates to warming of frozen samples over time. I would like to know how thick the styrofoam needs to be etc.

thanks

Answer 1

The hyperphysics site has a nice simple discussion of heat conduction:

http://hyperphysics.phy-astr.gsu.edu/Hba...

rather than restate it here, I'm going to refer you there for the heat conduction details.  

Hyperphysics also provides an estimate of the thermal conductivity of styrofoam:

http://hyperphysics.phy-astr.gsu.edu/Hba...

Now to answer your question, first thing is that without actively cooling the inside of your box, there is no way to keep it at -80 C using a finite thickness of styrofoam for any appreciable length of time.  What you will need to do is figure out how much warming you can tolerate over a specific time interval.  For example, let's say you have a 1 kg block of ice that is roughly 10 cm by 10 cm by 10 cm initially at -80 C and you can let that warm to -75 C over 10,000 seconds.  The heat capacity of ice is around 2 J/K-gm, which means you can let 2 J/K-cm * 1000 g * 5 K of heat, or 10,000 J into the box.  

By solving the formula on the Hyperphysics site, for a 5 cm thick layer of styrofoam I get a heat transfer rate of 2 W, or 2 J/s, which means in 10,000 s I'll only let 20,000 J into the box and my ice will heat above my specified temperature of -75 K over 10,000 s.  This means I need more insulation.  

How much you will need depends on what you have inside the box (because that determines the heat capacity) and precisely how much above -80 C it can rise in whatever time is relevant to you.  

There are also some simplifications I did which may or may not be relevant to you.  First, the rate you calculate is the initial rate, and since it depends on the temperature difference between the inside and outside of the box, as the box warms the heat transfer as a function of time decreases.  So the heating rate based on the initial heat flow is a minimum time, although in your case since the initial and final inside temperature is going to be small, the error in assuming the heating rate is constant will be small.

Of more importance is that whatever is inside your box won't heat evenly.  The surface will be much warmer, in general, than the core.  So in my example, what would happen is the surface of my ice might rise to -60 C, with the center of the block only dropping to -79 K.  If I integrate the total heat that went into the ice it will be close to the 20,000 J and the average temperature will be close to -70 K.  

But the bottom line is that -80 C (dry ice temperature) is cold and the heat flow will be large as a result.  In order to keep it around that temperature for a significant amount of time you are going to need a lot of styrofoam.