Keeping a reef tank warm with calcium chloride during a power outage?

[BigJohnny]

Hey Randy,

Apparently 100g of calcium chloride in a 160z water bottle would heat up to about 130*F pretty quickly without giving off any gas that would pop the bottle, which is pretty wild. Float a couple of those bad boys in your tank or even larger 1 gallon jugs and I think that would work pretty well. My question is, is there anyway to estimate how much calcium chloride/those water bottles I would need to maintain 78*F in my 90g display? Ambient temperature is 73*F and likely would stay there for the duration of any outage (hopefully). I understand there are many variables but just looking for a basic idea of what I would need so I can determine if I need to go run and get some more bottles/calcium chloride or not.

Thanks!

 

[Mark Derail]

Ya Joey DIY King on YTube did a vid on that, oh, maybe 5 yrs ago - if not more.

I buy it in bulk during the summer - 5g tubs - from the local swimming pool store, as well as swimming pool salt.
Mix 50-50 for the winter, and when you have ice storms, like the next day was it gets really cold and your stairs are treacherous - this stuff melts right through, even when it is -20c or -30c. Thanks to the calcium chloride.

 

[BigJohnny]

Ya Joey DIY King on YTube did a vid on that, oh, maybe 5 yrs ago - if not more.

I buy it in bulk during the summer - 5g tubs - from the local swimming pool store, as well as swimming pool salt.
Mix 50-50 for the winter, and when you have ice storms, like the next day was it gets really cold and your stairs are treacherous - this stuff melts right through, even when it is -20c or -30c. Thanks to the calcium chloride.

Yes I actually saw that video today that's how I found out about it. I knew it was the same stuff to melt ice but just never considered using it to keep an aquarium warm during a power outage.

 

[vegamedic]

This isn't an easy question to answer because most of the "easy" heat transfer formulas (Q = mcΔT, etc) assume a closed system, among other things. You'd be trying to heat a system with lots of things with different specific heats (glass, water, air, rock), continuing to lose heat to air, which is evaporating at a non-constant rate, with a heat source that's degrading over time...encapsulated in plastic. Not to mention you'd probably get some significant temperature stratification depending on the size of the tank.

edit: I did some work in this area for a program I wrote for my other hobby (homebrewing) as some brews require mash infusions to step the temp up. That was adding water directly to water in an insulated, closed system and still a PITA for the above reasons.

 

[BigJohnny]

This isn't an easy question to answer because most of the "easy" heat transfer formulas (Q = mcΔT, etc) assume a closed system, among other things. You'd be trying to heat a system with lots of things with different specific heats (glass, water, air, rock), continuing to lose heat to air, which is evaporating at a non-constant rate, with a heat source that's degrading over time...encapsulated in plastic. Not to mention you'd probably get some significant temperature stratification depending on the size of the tank.

edit: I did some work in this area for a program I wrote for my other hobby (homebrewing) as some brews require mash infusions to step the temp up. That was adding water directly to water in an insulated, closed system and still a PITA for the above reasons.

Yea I'd assume several assumptions would need to be made. Any rough estimates? Lol

 

[Mark Derail]

Of course @vegamedic is correct. Thermodynamics. It's why we don't have flamethrowers on snow clearing trucks in the winter.

Last time I had used up all my Calcium Chrolide (it was winter). But had lots of hot water available, so floated lots of plastic bottles filled with hot water.
I honestly raised the 50g DT from 25.3 to 25.4 then gave up after some 6x transfers.
I was using 24 plastic bottles, changing them every 20 mins or so.

Keeping the room warm will help a lot more. The hot water bottles at best maintained things for maybe an hour and helped reduce evaporation.
I pumped up the gas fireplace to max to get the room warm.
Lucky the power outtage was only a half day.

 

[BigJohnny]

Of course @vegamedic is correct. Thermodynamics. It's why we don't have flamethrowers on snow clearing trucks in the winter.

Last time I had used up all my Calcium Chrolide (it was winter). But had lots of hot water available, so floated lots of plastic bottles filled with hot water.
I honestly raised the 50g DT from 25.3 to 24.4 then gave up after some 6x transfers.
I was using 24 plastic bottles, changing them every 20 mins or so.

Keeping the room warm will help a lot more. The hot water bottles at best maintained things for maybe an hour and helped reduce evaporation.
I pumped up the gas fireplace to max to get the room warm.
Lucky the power outtage was only a half day.

Any little bit helps and depending on circumstances it can keep your tank warmer for a lot longer than an hour without replacing every 20 mins. For example if I wrapped the tank in newspaper and blankets and used several gallon jugs at 60* C. I'm just looking for some fuzzy math. Even calculations for a closed system would atleast let me know I'd need a lot more than that lol.

 

[Mark Derail]

I meant I raised 0.10c in an hour, plus maintaining it at a constant temp. My home hot water tank is a 60g, and I used up maybe half in my many trips.
The calcium chloride gets hotter though and the heat generated lasts longer.
Joey's vid had some numbers I believe.

 

[Randy Holmes-Farley]

The heat of hydration is known, so it is not hard to determine how much water can be heated this way, but the main uncertainty is the hydration level of the starting calcium chloride. The di hydrate will heat a lot less than the anhydrous form, and anhydrous will still have some moisture unless you make it yourself by ensuring you drive off all the moisture.

 

[BigJohnny]

The heat of hydration is known, so it is not hard to determine how much water can be heated this way, but the main uncertainty is the hydration level of the starting calcium chloride. The di hydrate will heat a lot less than the anhydrous form, and anhydrous will still have some moisture unless you make it yourself by ensuring you drive off all the moisture.

Currently I only have brs calcium chloride but can go pick up some driveway melt stuff at home depot, that's what joey used in diy video and got a water bottle from 25c to 55c with 100g

 

[Randy Holmes-Farley]

Anhydrous calcium chloride gives off about 81 kj/mole when it dissolves.

The specific heat of normal water (seawater won't be much different) is 4.2 kJ/kg K.

Thus, 4.2 kJ will raise 1 kg of water by 1 deg C.

Thus, one mole (111 grams) will raise 1 kg of water by 81/4.2 = 19 deg C.

 

[Randy Holmes-Farley]

The dihydrate gives off about 45 kj/mole when dissolving.

So 1 mole (147 grams) gives 45 kj of heat.

That raises 1 kg of water by 11 deg C.

1 kg of water is about 1 L.

The data for both of these calculations is given here:

https://www.sciencedirect.com/science/article/pii/0021961485900837

 

[BigJohnny]

Anhydrous calcium chloride gives off about 81 kj/mole when it dissolves.

The specific heat of normal water (seawater won't be much different) is 4.2 kJ/kg K.

Thus, 4.2 kJ will raise 1 kg of water by 1 deg C.

Thus, one mole (111 grams) will raise 1 kg of water by 81/4.2 = 19 deg C.

The dihydrate gives off about 45 kj/mole when dissolving.

So 1 mole (147 grams) gives 45 kj of heat.

That raises 1 kg of water by 11 deg C.

1 kg of water is about 1 L.

The data for both of these calculations is given here:

https://www.sciencedirect.com/science/article/pii/0021961485900837

Great thanks randy, just what I was looking for. So essentially I would need at a minimum 1753g of anhydrous calcium chloride to raise 80g/300L 1 degree C. So if the temp of my tank was dropping 1 degree C per hour I would need to add that much in bottles every hour, essentially (I realize not that simple)

Now for the 2nd part, how much can I concentrate it to use less water bottles and get the same increase?

 

[Randy Holmes-Farley]

Great thanks randy, just what I was looking for. So essentially I would need at a minimum 1753g of anhydrous calcium chloride to raise 80g/300L 1 degree C. So if the temp of my tank was dropping 1 degree C per hour I would need to add that much in bottles every hour, essentially (I realize not that simple)

Now for the 2nd part, how much can I concentrate it to use less water bottles and get the same increase?

Yes on part 1.

The amount of heat declines a bit with the concentration at the end of dissolving, but the calculation I did made no assumption on the concentration after dissolution (actually, meaning it is pretty dilute, which is the normal scientist assumption on things like this).

 

[BigJohnny]

Yes on part 1.

The amount of heat declines a bit with the concentration at the end of dissolving, but the calculation I did made no assumption on the concentration after dissolution (actually, meaning it is pretty dilute, which is the normal scientist assumption on things like this).

Great so if I wanted to use 4.2g of water bottles (80 gallons/19) of water to raise my 80g of tank water about 1*C I could use the 111g/L. You stated.

Do you think I could double that concentration so 111g/500ml without losing much heat? That is about what the king of DIY used in 100g/16 fl ounce water bottle

 

[Randy Holmes-Farley]

I'm confused. I didn't mention a concentration.

The comment that:

"So 1 mole (147 grams) gives 45 kj of heat.

That raises 1 kg of water by 11 deg C.

1 kg of water is about 1 L."

Did not mean to imply that the CaCl2 was dissolving in that liter (although I guess it could). It was intended to show how much the heat given off (by dissolving in some volume) would raise the temp of a liter.

 

[BigJohnny]

I'm confused. I didn't mention a concentration.

The comment that:

"So 1 mole (147 grams) gives 45 kj of heat.

That raises 1 kg of water by 11 deg C.

1 kg of water is about 1 L."

Did not mean to imply that the CaCl2 was dissolving in that liter (although I guess it could). It was intended to show how much the heat given off (by dissolving in some volume) would raise the temp of a liter.

I assumed that since you were explaining how much a given amount of CaCl2 dissolved in a liter would raise the temp of that liter that it would in fact dissolve in that liter, my bad lol. I get it now.

So independent of that example how much do you think I can dissolve per liter. Also my math was based of the 111g anhydrous not the 147 dihydrate

 

[BigJohnny]

Although I only have brs calcium chloride now, whatever that is.

Curious for anhydrous as well because I plan to stock some just in case

 

[Randy Holmes-Farley]

The solubility is very high (>500 g/L) and you can dissolve a lot, but the heat may be reduced a bit at higher concentrations.

 

[chipmunkofdoom2]

Randy, would calcium oxide be a better choice if the goal is to create heat in an emergency situation? It releases 64kj/mole upon hydration if I'm not mistaken. Plus, if my Googling is correct, calcium oxide weighs less than calcium chloride per mole. So a smaller amount of CaO would give a larger temperature increase as compared to CaCl2 (again, if my understanding is correct).