# Thermal expansion of water

1. Oct 3, 2015

### de_Sitter

Hello,

I'm trying to determine a way of calculating the thermal expansion of a volume of water. The formula I have come across is:

ΔV = βV0ΛT

The general consensus seems to be that water expands roughly 4% from 20°C to 100°C, or 4.2% from 4°C to 100°C (http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html). Using β = 0.000214 as found on (http://www.engineeringtoolbox.com/cubical-expansion-coefficients-d_1262.html) the result of this formula is not around the expected 4% from 20 - 100°C. e.g. 25L gives ≈ 0.43 L increase.

Am I using the formula correctly? Some rudimentary examples use β as a constant, while others remark that β varies with temperature, but don't provide examples of using this in practice. Any suggestions would be appreciated.

Thank you.

2. Oct 3, 2015

### Bystander

CRC Hndbk. tables of density of water. Definitely not constant for water.

3. Oct 3, 2015

### Staff: Mentor

The coefficient of volume expansion varies with temperature. The equation is only supposed to apply in the vicinity of the base temperature. In your case, the coefficient is for around 20 C.

Chet

4. Oct 3, 2015

### de_Sitter

Hello,

Thank you both for your replies. I'm a little unconvinced that using the coefficient of the base temperature is correct. Using 25L from 20 to 100°C with coefficient 0.000207 as per (http://www.engineeringtoolbox.com/volumetric-temperature-expansion-d_315.html) I get ΔV = 1.9715 L, which is around a 7.8% increase, rather than the expected 4%, as given on (http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html). Does anyone have an example of how they would model this increase?

5. Oct 3, 2015

### Staff: Mentor

To do it over a larger temperature range, you would have to know the thermal expansion coefficient as a function of temperature, and you would have to integrate.

6. Oct 3, 2015

### nasu

You don't get that value for that coefficient of thermal expansion assumed constant over the temperature range. What you get is around 0.4 L or about 1.6%.
But you have a table with the variation of the coefficient with temperature right there. You can see that it increases with the temperature so it will be reasonable to get 4% over the whole range.
You can do an estimate by using these values.

7. Oct 3, 2015

8. Oct 4, 2015

### de_Sitter

Thank you all for your assistance. I'm a little rusty on my calculus, but Nidum's links are perfect for my purposes.

Thanks again!

9. Oct 4, 2015

### de_Sitter

So just to confirm I'm doing it right Chestermiller. I've fitted a curve to the data points given in some of the tables I linked earlier, which gives β(T). I then find the definite integral of this function over the temperature range (e.g 20°-100°C), then multiply by the initial volume?

ΔV = ∫[β(T)] dT ⋅ Vo , with 20° and 100° as the limits of integration

Last edited: Oct 4, 2015
10. Oct 4, 2015

### Staff: Mentor

Basically correct for a small temperature change, but I would do it a little differently. The exact equation is:

$$\frac{1}{V}\frac{dV}{dT}=β(T)$$

So, integrating this equation, I get:

$$V=V_0e^{\int_{T_0}^T{β(T')dT'}}$$

where T' is a dummy variable of integration. This equation will give a more accurate result.

Chet

11. Oct 4, 2015

### de_Sitter

Thanks Chet, I'm not quite familiar with this dummy variable concept, but I'll do some reading. Thanks for the input.