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Rate of temperature change |
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| Jun24-12, 05:56 AM | #1 |
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Rate of temperature change
Hi all:
Sorry for the basic question but I have come across the formula ºC.min-1 in a paper which I believe is the formula for temperature change during one minute. The thing is that when I go to see how the authors used it, they have greater temperature change values than the ones a basic use of the formula implies in my interpretation (for instance, T1=27º; T2= 28º; T1 and T2 is one minute span so rate of temp change is 1ºC). Is that right or I'm missing something in the formula? So, basically I have a data series of temperatures (as the authors of the paper) taken every minute and I want to know what is the rate of temperature change. So, which formula (and how) to use it? Thanks! |
| Jun24-12, 06:52 AM | #2 |
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While it may be true that from 27 C to 28 C, the rate of temperature change per minute is 1 C, we have no idea how it would change at other temperatures! You gave no description, or even a graph, of the temperature change. It is entirely possible (an in fact, probable if one understands Newton's Cooling Law) that the rate of change of temperature may not be a constant. Zz. |
| Jun24-12, 08:32 AM | #3 |
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Thanks for the answer.
I attach the original graph I saw in the paper I mention and a graph with my data series to see if it helps but I believe that the rate of temperature change, in these cases, is not constant. So, what I basically want to do is to plot in my graph a rate of temperature change similar to what appears in the graph taken from the paper. Thanks! |
| Jun24-12, 03:44 PM | #4 |
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Rate of temperature changeeg it's not "1ºC" Basic calculation would be.. = (Temperature 2 - Temperature 1) / (Time 2 - Time 1) = (28-27)/1 = 1 ºC.min-1 Checking the units... = (ºC - ºC) / (min - min) = ºC/min = ºC.min-1 |
| Jun24-12, 04:02 PM | #5 |
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Are your times in mins and seconds or hours and mins? I've assumed mins and seconds below.. Looking at the left hand side of your graph... At time = 13:00.5, T = 26.3 At time = 13:01.5, T= 26.1 So at 13.01 the slope is =(26.1-26.3)/(1.5-0.5) =-0.2/1 =-0.2 ºC per second Negative because the temperature is falling. |
| Jun24-12, 06:40 PM | #6 |
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Thanks, my times are actually hours and minutes but it does not matter.
I thought so but if you look at the second graph (from the paper) you realise that they could not have used that formula since they plot, for instance, a 2.2 ºC.min-1 when in that minute (from 12:14 to 12:15) there is a very small temperature difference between the two times (0.1 ºC or 0.2 at most)... So how are they using the formula to give these results? Are their calculations accurate? I'm still missing something? Thanks |
| Jun25-12, 03:49 AM | #7 |
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You are correct.
There appears to be something wrong with their graph (or our understanding of it) because even when the temperature is falling the rate of change is shown as +ve. |
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