Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is it possible to predict what the final temperature will be by taking temperature samples. For example, an object is 0ºC and moved to a room above 0ºC. I'm taking temperature of the object using a thermometer every second. Can I predict (approximation) on what the final temperature would be after a few samples? I guess the more samples the more accurate it would be. Can I calculate when the final temperature might occur based the rate of the temperature change?

Is there any formulas for these kind of calculations?

share|improve this question
Seems certain that a 0 degree body moved into a 25 degree environment will ultimately reach a final temperature of 25 degrees. –  Michael Luciuk Feb 4 '13 at 16:49
The temperature is unknown by the thermometer. –  willi Feb 4 '13 at 16:55
add comment

1 Answer

up vote 2 down vote accepted

Assuming cooling is mainly by convection, the cooling will be described by Newton's law of cooling. This states that the rate of temperature change is proportional to the temperature difference, and result is that the difference between the temperature of your object and the room decays exponentially:

$$ T_{room} - T_{object} = (T_{room} - T_{0})e^{-kt} $$

where $T_{0}$ is the initial temperature of your object and $k$ is some constant.

You can take the temperature of the object as a function of time and then fit the expression above, but the problem is that this type of fit gives rather large errors in the final temperature $T_{room}$ unless you measure for long enough that you've almost reached the final temperature. You may also find Newton's law of cooling breaks down when the temperature difference is very small, because there is no longer effective convection.

share|improve this answer
+ I came up with that method in the context of pharmaco-kinetics, and a friend said it had a name: Aitken Acceleration. –  Mike Dunlavey Feb 4 '13 at 22:44
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.