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Sat, 16 Feb 2008

How to measure a temperature of the body under very limited conditions.

Let's suppose one does not have a thermometer, but there are lots of instruments and equipment around starting from screwdriver to drills, from simple amper/voltmeter to laptop. As a prompting: there is also electricity, vater and automatic teakettle.
Task is to measure temperature of the own body and decide to get or not to get an aspirin. Or make some fun from the process because of quite boring sickness.

Solution is pretty geeky, but first try to think about it yourself.


So, the solution.
It is based on the fact, that when human body or part of it is placed into environment with essentially the same temperature, but much bigger thermal capacity, it does not feel this. Try get shower with about 36 degress Centigrade, and you will not feel neither cold, nor hot. Things are different when air on the street is more than 30 degrees Centigrade, that is because of too much different thermal capacity of water (it is huge) and air (very small).

So, back to the task. To determine your tempeperature you have to get precise volume of water in the teakettle (let's say 1 liter, I could measure it because I have water counters), connect teakettle to the electricity via ampermeter, measure voltage by voltmeter.
Then you have to put your arm into the teakettle and turn it on (beware of heating element) and checkout first time. When your hand will feel itself very comfortable (here is a main error factor) you have to checkout second time. Then remove your arm and wait until water become boiled and write third time.

Now, its time for school physics: power of the teakettle, which is equal to multiplication of current strength and voltage, multipled by time difference is equal to weight of the water multipled by its thermal capacity and temperature difference, which was changed during above time frame.

So, here are practical results:
current strength I = 3.7 A
voltage V = 231 V
mass of the water m = 1 kg
thermal capacity of the water c = 4200 J/(kg*degree)
time difference for complete boiling (from unknown temperature to 100 degrees Centigrade) dt0 = 420 seconds
temperature difference dT can be found from following equation:
I*V*dt0 = m*c*dT
So, we have dT = 100 (temperature of the boiling) - T0 (initial temperature of the water) = I*V*dt0/(m*c), and is equal to 85 degress Centigrade, so initial temperature of the water was about 15 degress Centigrade.

Time difference between start of the process and comfortable temperature was about 30 seconds, so placing this timeframe into above equation we can find, that temperature was changed by 6 degress.

Since we already found, that initial temperature was about 15 degress Centigrade, calculated temperature of my body is about 21 degrees Centigrade.
Its time to go back to grave...

P.S. Yes, I'm a former looser-physicist, that's why I became a kernel hacker, this can explain alot...

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