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« Last post by **MJM460 ** on* ***Today** at 08:58:01 AM »
Hi Willy, I probably should have included the equation for that Newtonian cooling we were discussing yesterday, so here it is.

T(t) = T(a) + (T(0) - T(a)) x e^(-k x t)

As usual I should use subscripts, but on the iPad, it's too hard. Read it like this :-

The temperature at time t, T(t) equals the ambient temperature, T(a) plus the difference between the initial temperature, T(0), and T(a), or (T(0)-T(a)), raised to the power of (that ^ symbol), minus k x t.

Now this might need a bit more explanation. That k is a time constant (not thermal conductivity) and t is the elapsed time in the units you choose, seconds, minutes, hours or whatever is convenient. The units are not important, but will affect the value of k in your calculation, as it has units which must be based on the same time units. Also, raising a number to a negative power gives the reciprocal of raising it to a positive power. The symbol, e, is a well known standard mathematical constant, and is the base for natural logarithms, but that is another subject. Your calculator has it, as does any spreadsheet program.

So, using any two coincident time and temperature readings from your cooling readings, you can determine k with a little simple algebra. I used minutes and calculated k = 0.00496, using the times for 138 degrees and 43 degrees. I spent a bit of time looking at the value of k in terms of the insulation conductivity, the convection coefficient, density, surface area and so on, but decided it was more complex than it was worth. Unfortunately k is not totally a constant but to see how much it varies you would need a few more readings in the first 100 degrees of cooling to compare with the value I obtained and also to calculate more values from the last bit of cooling, a great job for a spreadsheet. The formula is a mathematical equation which closely enough follows the same curve as the actual cooling, and does indeed predict infinite time for the boiler to reach ambient temperature.

However, in practice two things happen that mean the temperature is reached sooner. First, even the equation soon predicts that the temperature difference is no longer large enough to be detectable. You can calculate how long to get to 19 degrees, the smallest temperature you can measure with your instrument above the ambient of 18. Second, and equally important is that particularly when you are talking about cooling times of several hours, the ambient temperature is likely to change, thus moving the cheese so to speak. However the time constant, k is useful in indicating how quickly the cooling is likely to occur, the very low value in this case is caused by the insulation and means a long time to cool. A high value would be found with no insulation and indicate a shorter time to reach equilibrium temperature.

When your steel and amber are in a room long enough to reach room temperature, as you say, they will all be at the same temperature as the room and almost everything else it the room, however your finger is not. Your blood supply is trying to keep the finger up to about 37, though it often does not quite make it in cold weather as we all know. So when you touch anything at 20 degrees, there is heat flow to the object controlled by the contact resistance of your finger, and the conductivity of the object. Your nerves are close to the skin, again every day experience, so your finger can feel itself being cooled by the steel which of course conducts very well. When you touch the amber, your finger is not cooled so much as the amber does not conduct so well. The situation is controlled by the steady state heat flow from your finger which is receiving heat from your blood, to the object you touch.

The thermocouple is quite different as there is no heat source. Heat only flows from the hot object it is placed against until the thermocouple reaches the temperature of the object it is touching. So that time constant does determine the time it takes the thermocouple to reach the temperature of the object, when the zeroth law of thermodynamics says there is no further heat flow.

There is a new one for you. It really is called that. While it was only clearly put into words after the first and second laws, logically it is necessary to have this one before you can properly deduce the others. And it cannot be deduced from other laws. Definitely still thermodynamics. If you put some insulation outside the thermocouple so there is minimal loss to the outside air, and use it to hold the junction, or the sheath in the case of your instrument, in close contact with the object you want to measure, you will get both better accuracy and a quicker response to temperature changes. Oh, and wrap the wires (or the part of the sheath, in contact with the air with some insulating material to minimise heat flow along the wires. This dependence on conduction to make the temperatures equal is the reason temperature measurement always has a slower response than pressure.

It looks like consideration of engines will have to wait for another day,

Thanks for looking in,

MJM460