Buying hundreds of pounds of ice is such a hassle. For years I’ve scouted the cheapest places to purchase ice, and then hauled it around in a massive cooler to make it through a night of demo derby. Now when I was in school I took a thermodynamics class, and its not all that hard to lay out the thermal capacity of any particular volume of water. This post will show why we can all save some money on ice…I sure do enjoy saving money!

So to start we have to make some assumptions. I’ll assume the size of the tranny cooler to be built around a 20mm ammo can. I’ve seen many coolers built with these, and its about the smallest volume I would want to use. Please note that in larger size coolers the additional water volume will add further thermal capacity. These boxes are 17″ long, 7″ wide, and 15″ tall, which makes for about 7.7 gallons of water capacity. For this instance, I will not attempt to assume a volume of the heat exchanger, be it a finned unit or a coil of copper, but it will take up some volume, so to make a round number lets assume 7.5 gallon capacity with heat exchanger. From here, I’ll go through two scenarios of thermal capacity, one packed full of ice, and one at 32°F water.

*Side Gripe. My brother has an icebox tranny cooler he purchased from a derby vendor several years ago, and it has a super tough plate frame heat exchanger, but it is mounted midway in the box, so that after half the water is evaporated the exchanger is above the water level. In order for the exchanger to transfer heat to the water it must first be immersed in it, so the heat exchanger should be mounted near bottom. That seems intuitive to me, but shame on that vendor for a poor design. FYI: I don’t remember the name of the vendor, so I don’t feel bad throwing them under the bus, and my brother has never taken the time to fix it either.

Scenario 1 – Full of Ice

Conveniently, 7.5 gallons is almost exactly 1 cubic foot. The density of ice is 57.371 lbs/ft^3, so if you were to have a solid block of ice in this tranny cooler, the max you could get would be 57.5 lbs. I will assume the ice to be at 32°F. The latent heat of melting is what we need to figure out how much heat the ice will absorb before it melts entirely. This is 143.4Btu/lb, and there is 57.5lbs, so all we have to due is multiply these together and come up with **8,245** Btus of heat absorbed before ice is melted and then we have 57.5lbs of water at 32°F (Not very impressive at all). Next, the water has to be heated to its boiling point, and we need its specific heat capacity 1.798 Btu/lb per degree change. Now we just multiply 57.5lbs by 1.798Btu/lb per degree and multiply again by 180° (212°F-32°F) and get **18,606**. This is more than double the energy needed to melt it from ice. Then there is the latent heat of vaporization, which is the additional energy that it then takes to vaporize all that 212° water. This one is a simple 970Btu/lb, and multiply by 57.5lbs to get **55,794 **Btus of thermal capacity. Notice how much more energy it takes to vaporize the water than to melt it…. That makes a total ideal thermal capacity of this tranny cooler with an ice block **82,645 Btus**.

Scenario 2- Full of 32 degree water

You might think I could start my calculations on the water with the numbers figured above, and while it would be close, water and ice do not have the same density, so 7.5 gallons of water weighs more than 7.5 gallons of ice. The density of water is 62.4 lbs/ft^3, so 5 more pounds of water in our cooler in this scenario. Then we just need to recalculate the specific heat and vaporization capacities. *Specific Heat – *62.5lbs X 1.798 Btu/lb X 180° = **20,228 Btus**. *Latent Heat of Vaporization –*62.5lbs X 970 Btu/lb = **60,625 Btus**. That makes to the total ideal thermal capacity of this scenario **80,853 Btus**.

The full of ice scenario only beats the cold water by 1,792 Btus! Also, I’ve yet to see anyone remove their tranny cooler from the car and freeze it entirely, so most likely the actual numbers would fall somewhere in between. ** My main take away is that the additional heat capacity of the ice is almost completely negated by the extra volume the ice takes up over water. **

From a practical standpoint what makes sense to me to proceed with is to fill up the cooler mostly with water (assuming 50° from a hydrant), put in two 10lb bags of ice, and top it off with water. *62.5lbs*1.798btu/lb per degree x 18° = 2,023 btus; 2,023btus / 143.4 btu/lb = 14.1lbs ice.* You can do the math on how much ice is needed to cool a volume of water at any temperature to 32° like I did here, or leave a comment, and myself or someone may be able to help you.

If this gets your brain wondering things like, “Hmm, I wonder how many btu’s per hour a th400 produces” or “I wonder what the ideal temperature is for ATF anyhow?” or “Does a coil of copper tubing work as well as a plate frame heat exchanger?” or “should I switch to an air cooled transmission cooler and avoid ice altogether?”, then perhaps I can look to address these quandaries in a future post. Ta Ta for now.