# How can I calibrate a limited-range thermometer?

I've got an electronic weather thermometer with a limited range (roughly -20°C to 40°C, in 0.1° intervals). How could I go about calibrating it using household equipment?

Getting a 0°C data point is easy: place it in a waterproof container in an ice bath. What other data points can I use? The traditional second point, boiling water, isn't available due to the limited range of the thermometer.

• As a future reference, please be aware of the following trivial curiosity: the correct way to give the temperatures in Celcius does not involve the "$\, ^\circ\,$" symbol, i.e. your values should be $-20$C to $40$C etc, and spoken as "twenty C" rather than "twenty degrees C". The same holds for the Kelvin scale. Only the Farenheit scale uses "$\, ^\circ\,$" and values are spoken in degree, i.e – ZeroTheHero Mar 1 '17 at 2:46
• @ZeroTheHero The unit of Celsius temperature is the "degree Celsius" with the symbol "$^\circ C$". physics.nist.gov/cuu/Units/units.html – Farcher Mar 1 '17 at 7:31
• Which type of sensor is used in your electronic weather thermometer? – Farcher Mar 1 '17 at 7:37
• @Farcher - I was about to ask that too, but concluded it doesnt matter to answer the specific question (ie. a second reference temperature within the measurement range)? Given the high precision and narrow range, I suspect its a thermistor. – theNamesCross Mar 1 '17 at 8:19

You could use a salt water binary. The thing is, salt water freeze below 0°C if there is not too much salt :

Note that in the diagram, x is the mole fraction of salt. (I took it from the subject of an exam, here, in french (pdf)). Below the horizontal line, the mixture is solid. Above the curved line, the mixture is liquid. More on binary diagrams here.

So what you could do is look up in a IUPAC book a mole fraction value of salt corresponding to a freezing temperature of say -15°C. Then you prepare the mixture with the help of your kitchen scale (I would recommend to use large amount of water and salt to minimize the relative uncertainties) and take half of the mixture to the freezer. When it's frozen, mix the frozen part and the liquid part, stir the whole thing until all the ice has melted and here you go, you have your second calibrating point (which is the point of the curved line corresponding to the mole fraction you choosed).

And you only used household equipment (and I assumed you don't have another thermometer).

Use the boiling point of another liquid. Example: a Google search indicates that diethyl ether has a normal boiling point of 34.7 deg. C. If you are sure that your ambient atmospheric pressure is 1 atm (e.g., at sea level), use that value. If your elevation is higher, look up the Antoine equation for this substance, fix the pressure to your ambient conditions, and solve the equation for temperature.

Note - diethyl ether is VERY flammable. Use a small amount of boiling diethyl ether in a hot water bath (NO OPEN FLAMES), and do your calibration outdoors with some amount of air circulation. Also, keep a water hose nearby.

If a two point calibration is sufficient, I recommend using the melting point of gallium as the second reference. Commonly used as a calibration reference, gallium is a readily available metal with a melting point at $29.7646 \text{[C]}$ - solid and liquid phases are mixed (like an ice bath) to achieve temporary equilibrium at the melting/freezing point. This is more accurate than using the boiling point of various liquids because they depend on atmospheric pressure, whereas the solid and liquid phases of gallium are (for practical purposes) independent of atmospheric pressure.

Beware that a two point calibration requires the measured temperature, bias, and uncertainty to be linear throughout the reference range. Similarly, measurements outside the calibration range should be treated with caution.

$$T_{meas} = \frac{T_{meas,2} - T_{meas,1}}{T_{ref,2} - T_{ref,1}}T_{ref} + T_{meas,1} \\ bias(T_{ref}) = \frac{bias(T_{ref,2}) - bias(T_{ref,1})}{T_{ref,2} - T_{ref,1}}T_{ref} + bias(T_{ref,1}) \\ u(T_{ref}) = \frac{u(T_{ref,2}) - u(T_{ref,1})}{T_{ref,2} - T_{ref,1}}T_{ref} + u(T_{ref,1}) \\$$

To account for possible non-linearity in the thermometer, additional calibration points are needed. Again, known material temperature references could be used. Or, another calibrated thermometer could be used to 'reference' your thermometer at chosen temperatures. Multi-point calibrations are beyond the scope of the question, but worth study if high precision is required.