What is the correct form of heat diffusion equation taking into account temperature dependence of specific heat 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
$$ \rho c\frac{dT}{dt} = \frac{d}{dx}\bigg(k\frac{dT}{dx}\bigg)$$
 A: 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 $\dfrac{\kappa}{c_v}$. 
A: It always will help you to go back to the derivation of the heat equation. The heat in a region $D$ at a time $t$ is given by the integral,
$$
H(t)=\int_Dc(t)\rho T(x,t)\,dx
$$
where we've added your assumption of $c$ being time-dependent.
Thus, the change in the heat would be
$$
\frac{dH(t)}{dt}=\frac{d}{dt}\int_Dc(t)\rho T(x,t)\,dt\equiv\int_D\rho\frac{\partial}{\partial t}cT\,dx
$$
Matching this with Fourier's law would give us something similar to your first equation (the difference being the use of partial derivatives, rather than total derivatives):
$$
\rho\frac{\partial}{\partial t}cT=\kappa\nabla^2T
$$
A: This is an old question, yet, it seems the answer is incorrect as it has little to do with the question.
Short answer: your first equation takes into account dependence of heat capacity on temperature, the second one does not.
Longer answer: Derivation of heat equation from energy balance (and making use of the Fourier law) leads to the following differential equation:
$\rho(t,x)f(t,x) = \frac{\partial}{\partial t}\left[\rho(t,x)c(t,x)T(t,x)\right] - \nabla\cdot\left[\alpha(t,x)\nabla T(t,x)\right]$,
where $\rho(t,x)$ is material density, $f(t,x)$ is heat generation within the material (e.g. radioactive decay), $c(t,x)$ is heat capacity, $\alpha(t,x)$ is heat conductivity. Now, for well behaved functions, we can write $c(t,x) = c(T(t,x))$, $\alpha(t,x)=\alpha(T(t,x))$. To explicitly write out the derivatives, use the chain rule (e.g. $\frac{\partial c}{\partial x} = \frac{\partial c}{\partial T}\frac{\partial T}{\partial x}$). The second equation comes from the assumption that the parameters are temperature independent, but both capacity and conductivity are temperature dependent but as temperature is space-time dependent, the are too space-time dependent.
