Fourier's law states that
"The time rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area."
A mathematical description of this law is given as
$$\frac{dQ}{dt}=-KA \frac{dT}{dL} \, . \tag{1}$$
where $K$ is the thermal conductivity of the substance in question, $A$ is the area of the substance normal to the direction of flow of heat current and $L$ is the lateral length of the substance.
However, we may write $dQ/dt=mC(dT/dt)$ -where $m$ is the mass of the substance and $C$ is the specific heat capacity of the substance-to obtain
$$m C \frac{dT}{dt} = K A \frac{dT}{dL} \, . \tag{2}$$
This would allow me to cancel the $dT$ on both sides of the equation to obtain
$$\frac{mC}{dt} = \frac{K A}{dL} \, , \tag{3}$$
which suggests that the time taken for the heat transfer to complete, between any two temperatures is a constant given by
$$ dt = \frac{dL m C}{KA} \, . \tag{4}$$
However, this is not at all the case.
What mistakes have I made in writing the above steps?
Best Answer
When you use the equation $\dot Q = - kA \dfrac {dT(x)}{dx}$ one of the assumptions which has been made is that you have steady state conditions.
That is the temperature $T$ only depends on position $x$ and not on the time $t$.
What you are trying to deal with is a situation where the temperature at a position $x$ also depends on time so the equation to use looks similar
$\dot Q = - kA \dfrac {\partial T(x,t)}{\partial x}$
but notice the subtle difference in that because the temperature depends on two variables, partial derivative need to be used.
This derivation shows you the sort of equation which needs to be used in such circumstances.
Imagine a slab of cross sectional area $A$ and thickness $\Delta x$ with the temperature on one side being $T(x)$ and on the other side $T(x+\Delta x)$ as shown in the diagram below
Unlike steady state conductivity problems here the amount of heat entering the slab $\dot Q(x)$ differs from the amount of heat leaving the slab $\dot Q(x+\Delta x)$, that difference being responsible for the temperature of the slab changing with time.
So the equation which balances the flow energy is
$\dot Q(x) - \dot Q(x+\Delta x) = \dfrac{ A \;\Delta x \; \rho\; c \;[T(t+ \Delta t - T(t)]}{\Delta t}$
where $\rho$ is the density of the slab and $c$ is the specific heat capacity of the slab.
Rearranging this gives
$- \dfrac 1 A \dfrac {\dot Q(x+\Delta x)-\dot Q(x)}{\Delta x} = \rho \;c\; \dfrac{T(t+ \Delta t - T(t)}{\Delta t}$
Now allowing $\Delta x$ and $\Delta t$ to tend to zero gives
$- \dfrac 1 A \dfrac {\partial \dot Q}{\partial x} = \rho \;c\; \dfrac{\partial T}{\partial t} \Rightarrow - \dfrac 1 A \dfrac {\partial }{\partial x} {\left( -k\;A\; \dfrac{\partial T}{\partial x}\right) }= \rho \;c\; \dfrac{\partial T}{\partial t} \Rightarrow k \;\dfrac {\partial^2 T(x,t)}{\partial x^2}= \rho \;c\; \dfrac{\partial T(x,t)}{\partial t} $
assuming that the thermal conductivity $k$ is constant.