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

I have the heat balance equation for a cooling case in the form:



  • $U$: energy (J)
  • $T$: temperature (K)
  • $t$: time (s)
  • $J$: heat flux leaving the system (W)

I have $U$ from experimental data; $U$ is a function of sample temperature, $T$, measured in the experiment, both are functions of the experiment time $t$.

Now I want to solve the differential equation for $T(t)$; but the problem is that $U(T)$ is not definite because the sample releases crystallization heat and causes $U(T)$ to increase after initial decrease.

I was thinking of using only $U(t)$ because this function would be definite. But I need $T(t)$ at the end.

Do I need to go in 3D or are there other suggestions?

share|cite|improve this question

Your question cannot be answered in its present form. The RHS is simply the heat flux out of the system, and that will be different depending on what your system is surrounded by and how those surroundings respond to having the temperature of their boundary raised.

A rough solution is afforded by something like Newton's law of cooling: this describes the $J$ as being proportional to the temperature difference between the body and the "ambient" (i.e. many diffusion lengths into the surrounding material). So, if your heat content is linear over your temperature range of interest, so $U = \sigma T$, where $\sigma$ is your system's heat capacity, and if $k$ is the constant in Newton's law of cooling then:

$$\sigma \,{\rm d}_t \Delta U = - k\, \Delta U$$

where $\Delta U$ is the temperature difference between your body and the "ambient", whence $\Delta U(t) = \Delta U(0) \exp(-k\,t/\,\sigma)$. You would put your experimental data in for $U(T)$ instead of my simple $\sigma\,T$ into the above differential equation and solve accordingly. $k$ is highly dependent on what surrounds your body. Some insulating materials / boundary conditions will keep $k$ very small, maybe even effectively nought depending on what timescales / temperature scales you want your simulation to be good for. Other materials are excellent heat conductors and will give a very high $k$. So you see you need to know a great deal more about the surroundings of your body to make headway with your analysis.

Another level of sophistication is to think of the body and surroundings not as "lumped" elements each with a uniform temperature throughout each of them and account for diffustion. Then your equation becomes a partial DE for a scalar temperature field:

$$\nabla\cdot \vec{J}(\vec{r},t) = -\partial_t U(\vec{r},t)$$

You get this equation by replacing the LHS of your equation with the volume integral of $U(\vec{r},t)$ where $U(\vec{r},t)$ is a scalar field describing the heat content per unit volume. Then the time rate of change of this quantity is the surface integral $\oint_S \vec{J} \cdot \hat{\vec{n}} \,{\rm d}S$ of the heat flux in through the surface at the boundary of the volume. The convert the surface integral to a volume integral by Gauss's divergence threorem. I.e. here is where you would put in your experimental data $U(T(\vec{r},t))$. The "localised" Newton's law of cooling is then $ \vec{J} = -\kappa \nabla T(\vec{r},t)$ where $\kappa$ is the materials heat conductivity and, if you approximate $U(\vec{r},t) = \rho\,\sigma\,T(\vec{r},t)$ where now $\sigma$ is the mass-specific heat capacity and $\rho$ the density, we get the diffusion equation for homogeneous mediums:

$$\kappa \nabla^2 T - \sigma\,\rho\, T = 0$$

which is the diffusion equation. You then solve this equation for two mediums: your object, embedded in an infinite other representing the surroundings. The boundary conditions are that $T$ and the normal to surface component of $\kappa \nabla T$ must be continuous across the boundary (the latter is just conservation of heat flux across the boundary).

share|cite|improve this answer

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.