What is the correct form of the heat diffusion equation in 1D if we take into account the temperature dependency of specific heat capacity?
$$ \rho\frac{d(cT)}{dt} = \frac{d}{dx}\bigg(k\frac{dT}{dx}\bigg)$$
or
$$ \rho c\frac{dT}{dt} = \frac{d}{dx}\bigg(k\frac{dT}{dx}\bigg)$$
Best Answer
The differential equation for the conduction of heat is: $$\mathbf{h} = -\kappa\mathbf{\nabla} T$$( This relationship is an approximate one, but holds good for many substances). Also, the equation of continuity for local conservation of heat flow is: $$ - \dfrac{dq}{dt} = \nabla\mathbf{h} \implies \dfrac{dq}{dt} = \kappa{\nabla}^2 T$$ where $q$ the amount of heat in a unit volume & $$\mathbf{\nabla}\cdot\mathbf{\nabla} = {\nabla}^2 = \text{Laplacian operator}$$ Now, we'll assume that the temperature of the material is proportional to the heat content per unit volume - that is, the body has a definite specific heat. So, we can write $$\Delta q = c_v\Delta T \implies \dfrac{dq}{dt} = c_v \dfrac{dT}{dt}$$. The rate of change of heat is proportional to the rate of change of temperature. The constant of proportionality $c_v$ is thd specific heat per unit volume of the material. Using this, we get $$\dfrac{dT}{dt} = \dfrac{\kappa}{c_v} {\nabla}^2 T$$. We find the time rate of change of $T$ at every point as proportional to Laplacian of T. We have a differential equation now for the temperature $T$ using specific heat. So the final equation is $$\dfrac{dT}{dt} = D{\nabla}^2 T$$, where $D$ is the diffusion constant , & is equal to $\dfrac{\kappa}{c_v}$.