Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I had a semi infinite, 1-D object and a finite 1-D object, both heated at the same constant rate at one end each for the same time period and both begin at the same initial temperature, is it physically meaningful for me to integrate along the length of the object and consider this integral as a function of time and a measure of the 'amount' of heat on the object?

share|cite|improve this question
up vote 1 down vote accepted

Temperature is the measurement of kinetic energy per unit particle mass. Since you've added the same amount of heat energy to each object, the finite object will have a higher temperature because its heat energy is distributed across a smaller collection of mass. Taking something's temperature is indeed a meaningful measurement ;)

share|cite|improve this answer
ah, so is it not meaningful to talk about the temperature at some point? Rather, we should talk about the temperature in some region? – user27182 Apr 24 '13 at 15:11
Is there no 'conservation of temperature' which means that the two objects should have the same total temperature? Can I calculate the internal energy of the objects and then get a conservation statement? – user27182 Apr 24 '13 at 15:13
You can absolutely take the temperature of something at a specific point - measure the average particle kinetic energy at the region you're interested in. Temperature is not a fundamental physical property in itself - it is in fact a measurement of the physical quantity, energy (which is conserved) within a given mass (also conserved :P). The two objects have had the same energy added to them, but their masses are different, so their temperature is not the same. It's important to note that the type of mass matters, when it comes to temperature. See – NWard Apr 24 '13 at 15:20

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.