Calcium Chloride not getting hot

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vadnad

vadnad

Member
Jun 17, 2019
20
St. Petersburg, FL
Pool Size
14000
Surface
Plaster
Chlorine
Salt Water Generator
SWG Type
Hayward Aqua Rite (T-9)
Anyone ever had calcium hardner not get hot? I just opened a new bag and mixed it with pool water in the same proportion i usually do, 1lb with half of a 5 gallon bucket and it didn't warm up. Also, typically after i stir for about 5 minutes the water is cloudy and this batch was clear. Any ideas? The Calcium Chloride looks like normal - white flakes. Pool water is 79F.
 
Last edited:
What brand calcium hardener are you using?

Show us pictures of the calcium hardener container.
 
To estimate the temperature increase when dissolving 1 lb of calcium chloride (CaCl₂) in 2.5 gallons of water, follow these steps:

Step 1: Determine the Heat Released​


The heat of dissolution (enthalpy of solution) for CaCl₂ is approximately -17.3 kcal per mole (exothermic, meaning heat is released).

  • Molar mass of CaCl₂ = 110.98 g/mol
  • 1 lb = 453.59 g, so moles of CaCl₂:
    453.59 g110.98 g/mol≈4.09 moles\frac{453.59 \text{ g}}{110.98 \text{ g/mol}} \approx 4.09 \text{ moles}
  • Heat released:
    4.09×17.3 kcal=70.7 kcal4.09 \times 17.3 \text{ kcal} = 70.7 \text{ kcal}
    (which is 70,700 cal)

Step 2: Determine the Water Mass​

  • 1 gallon of water = 8.34 lbs
  • 2.5 gallons =
    2.5×8.34=20.85 lbs2.5 \times 8.34 = 20.85 \text{ lbs}
  • Convert to grams:
    20.85×453.59=9,452 g20.85 \times 453.59 = 9,452 \text{ g}

Step 3: Calculate Temperature Increase​

Using the heat equation:

q=mcΔTq = mc\Delta T
where:
  • qq = 70,700 cal (heat released)
  • mm = 9,452 g (mass of water)
  • cc = 1 cal/g·°C (specific heat of water)
  • ΔT\Delta T = temperature increase

Solving for ΔT\Delta T:

ΔT=70,700 cal9,452 g×1 cal/g°C\Delta T = \frac{70,700 \text{ cal}}{9,452 \text{ g} \times 1 \text{ cal/g°C}}ΔT≈7.5°C\Delta T \approx 7.5°C

Final Answer:​

The water temperature would increase by about 7.5°C (13.5°F) under ideal conditions, assuming no heat loss to the surroundings.
 
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Take a test sample from your bucket and run a CH test on it.

Compare that to your fill water CH level.
 
When calcium chloride dissolves in water, it releases heat (an exothermic process). I'll calculate the temperature increase for your scenario.

To determine the temperature rise:
1. Convert units to a consistent system
2. Find the heat released when CaCl₂ dissolves
3. Calculate how much this heat will raise the water temperature

Step 1: Converting units
- 1 lb CaCl₂ = 453.59 g
- 2 gallons water = 7.57 liters = 7570 g
- Water temperature = 70°F = 21.1°C

Step 2: Heat released
The enthalpy of solution (heat released) for anhydrous CaCl₂ is approximately 81.3 kJ/mol.
- Molar mass of CaCl₂ = 110.98 g/mol
- Moles of CaCl₂ = 453.59 g ÷ 110.98 g/mol = 4.09 mol
- Total heat released = 4.09 mol × 81.3 kJ/mol = 332.5 kJ

Step 3: Temperature increase
The specific heat capacity of water is 4.18 J/(g·°C)
- Temperature change = Heat released ÷ (mass of water × specific heat capacity)
- Temperature change = 332,500 J ÷ (7570 g × 4.18 J/(g·°C))
- Temperature change = 332,500 J ÷ 31,642.6 J/°C = 10.5°C = 18.9°F

Therefore, you can expect a temperature increase of approximately 19°F, bringing the final temperature to about 89°F (32°C).

Note: This calculation assumes all the heat goes into warming the water and none is lost to the environment. In practice, some heat will be lost, so the actual temperature rise might be slightly less.
 
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vadnad said:
Good idea.
Click to expand...
So the results were that i tested a sample from my bucket and i stopped adding reagent after 34 drops when it didn't turn blue. That indicated the CH was high. I then tested my pool and it was 225. So i guess the CH is affective but maybe it's weak and not getting hot
 

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vadnad said:
So I guess the CH is affective but maybe it's weak and not getting hot
Click to expand...
If you dissolve 1 lb of calcium chloride anhydrous in 2.5 gallons of water at 70 degrees F, the water temperature be once the calcium chloride has completely dissolved will be about 85 degrees.

Are you using a thermometer to measure the temperature difference?

85 degrees is not really hot, so maybe you are expecting more heat than is produced?
 
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Calcium Chloride comes in different percentages, the 94% plus is usually for snow melt- I used the DOW branded Calcium Chloride pellets and they make the HD orange bucket very hot. I mix it with a drill paddle and let it cool before adding to the pool.

Dow Brand 94% Pellets are available at Janitorial supply houses for about $19 dollars for 50 lbs.
 
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Jsf721 said:
Calcium Chloride comes in different percentages, the 94% plus is usually for snow melt- I used the DOW branded Calcium Chloride pellets and they make the HD orange bucket very hot. I mix it with a drill paddle and let it cool before adding to the pool.

Dow Brand 94% Pellets are available at Janitorial supply houses for about $19 dollars for 50 lbs.
Click to expand...
I didn't think that it may be a different percentage than I've used before. Thanks.
 
Calcium chloride dihydrate (CaCl2·2H2O) is a common calcium salt and hydrate used in various applications, including as a calcium source, a catalyst, in food processing, and for ice and dust control.
Chemical Formula: CaCl₂·2H₂O
Appearance: White powder or flakes
Molecular Weight: 147.02 g/mol
CAS Number: 10035-04-8

Calcium chloride › Molar mass
110.98 g/mol.

Calcium chloride can be anhydrous or dihydrate.

The dihydrate is about 76% calcium chloride and 24% water.

Calcium chloride dihydrate (CaCl2·2H2O) contains approximately 24.49% water by mass.
Here's a breakdown of the calculation:
Molar mass of CaCl2: Ca (40.08) + 2 * Cl (35.45) = 111.08 g/mol
Molar mass of 2H2O: 2 * (2 * H (1.01) + O (16.00)) = 36.02 g/mol
Molar mass of CaCl2·2H2O: 111.08 + 36.02 = 147.10 g/mol
Mass of water in the hydrate: 36.02 g/mol
Percentage of water: (36.02 / 147.10) * 100% = 24.49%
 
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Last edited:
We have 1 lb of CaCl₂·2H₂O
Heat released = 1 lb × 131.1 BTU/lb = 131.1 BTU
Water amount = 2.5 gallons = 20.85 lb (1 gallon ≈ 8.34 lb)
Temperature change = Heat released ÷ (mass of water × specific heat capacity)

Water's specific heat capacity is 1 BTU/(lb·°F)
Temperature change = 131.1 BTU ÷ (20.85 lb × 1 BTU/(lb·°F))
= 131.1 BTU ÷ 20.85 BTU/°F
= 6.29°F
Therefore, the final temperature will be:
70°F + 6.29°F = 76.3°F
________________________________________________________

Amount: 1 lb of CaCl₂·H₂O
Heat released = 1 lb × 174.3 BTU/lb = 174.3 BTU
Water amount = 2.5 gallons = 20.85 lb
Temperature change = Heat released ÷ (mass of water × specific heat capacity)

Using water's specific heat capacity of 1 BTU/(lb·°F):
Temperature change = 174.3 BTU ÷ (20.85 lb × 1 BTU/(lb·°F))
= 174.3 BTU ÷ 20.85 BTU/°F
= 8.36°F
Therefore, the final temperature will be:
70°F + 8.36°F = 78.4°F
_______________________________________________________________________
Amount: 1 lb of anhydrous CaCl₂
Heat released = 1 lb × 317.2 BTU/lb = 317.2 BTU
Water amount = 2.5 gallons = 20.85 lb
Temperature change = Heat released ÷ (mass of water × specific heat capacity)

Using water's specific heat capacity of 1 BTU/(lb·°F):
Temperature change = 317.2 BTU ÷ (20.85 lb × 1 BTU/(lb·°F))
= 317.2 BTU ÷ 20.85 BTU/°F
= 15.21°F
Therefore, the final temperature will be:
70°F + 15.21°F = 85.2°F
 
The maximum solubility of calcium chloride anhydrous is about 0.745 lb per 1 lb of water.

To calculate the temperature rise, find the heat of solution and multiply it by the number of lbs.

For example, 0.745 lb calcium chloride dissolved in 1 lb of water will produce 0.745 x 317.2 = 236.3 btu of heat released.

For 1 lb of water, this is an increase of 236.3 degrees F.

For 10 lbs of water, it is 23.63 degrees F.

1 lb CaCl2 in 1 gallon of water (8.34 lb) = 317.2/8.34 = 38 degrees temperature rise.

As long as you start with cool to warm water and limit the calcium chloride to 1 lb anhydrous to 2 gallons of water, the temperature rise (19 degrees) should not be enough to make the water dangerously hot.

If you start with a water temperature of 91 degrees and you get a 19 degree temperature rise, then that is a final temperature of about 110 degrees.

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