What is the difference between temperature difference and temperature change?

Question

In a course of mathematical modelling that I am taking, there is a great confusion between the concepts of temperature change due to a unit heat input at some position $x$ and time $t$, and the temperature distribution due to a unit heat input at some position $x$ and time $t$.

The instantaneous plane source solution is: $$T\left ( x,t \right )=\frac{Q'}{2\sqrt{\pi \kappa t}}e^{-x^{2}/4\kappa t}$$ Now, my book attempts to connect the idea of temperature distribution. It denotes the temperature distribution due to the release of a unit of heat per unit area at $x=0$ and $t=0$ in an infinite region by $G(x,t)$: $$G\left ( x,t \right )= \begin{cases} 0, &\forall t< 0 \\ \frac{\mu}{\sqrt{t}}e^{-x^{2}/4\kappa t},& \forall t> 0\end{cases}$$ where $$\mu= \frac{1}{2\rho c\sqrt{\pi \kappa}}= \frac{1}{2\sqrt{\pi \kappa c \rho}} $$

Would someone be nice enough to explain to me the idea of temperature distribution and how it relates to temperature change?

Is temperature distribution $G(x,t)=\frac{T(x,t)}{Q'}$?

Edit: Thanks for cleaning up.

Answer 1

Your first expression is a temperature distribution: it simply tells you the temperature $T$ at each point in the $x,t$ space.

Now if you, using that formula, calculated two temperatures $T_1$ (at $x_1,t_1)$ and $T_2$ (at $x_2,t_2)$, then the difference:

$$\Delta T=T_2-T_1,$$

is an actual temperature difference. An infinitesimal temperature difference would be written as $\partial T$.

You'll probably need to use infinitesimal temperature differences to calculate how your temperature distribution evolves in time because you indicate there's heat input. Fourier's Law of heat conduction:

$$\large{q=-\kappa \frac{\partial T}{\partial x}},$$

where $q$ is the heat flux and $\frac{\partial T}{\partial x}$ a temperature gradient (ratio of two infinitesimal differences), is usually needed to solve that kind of problem. What Fourier really means is that heat conduction is favoured by greater temperature gradients.

Edit: (in answer to OP's comment)

We can re-write OP's first equation as:

$T(x,t)=Q'f(x,t)$

Dividing both sides by $Q'$ we get using OP's notation:

$G(x,t)=f(x,t),$ where $f(x,t)$:

$f\left ( x,t \right )=\dfrac{e^{-x^{2}/4\kappa t}}{2\sqrt{\pi \kappa t}}.$

By definition $G(x,t)$ cannot be a temperature distribution, although it clearly is a (Gaussian) distribution in $x,t$. I suspect the author will want to use this distribution further down in his derivation.

You can also work out the unit of measurement of $G(x,t)$: it is not $\mathrm{K}$, the unit of temperature.