TL; DR The material with the greater effusivity will be more likely to burn you upon contact.
Analysis of a simplified case
First consider the case where your palm comes into contact constant temperature wall. Often, we can consider your palm as a semi-infinite solid. The requirement for the semi-infinite approximation is that $T(x \rightarrow \infty, t) = T_i$. In other words, the temperature far away from the contact point ($x=0$) at any time $t>0$ is not affected by contact with the solid. This is a reasonable starting point for answering this question.
Assumptions:
1. Both the palm and the object are semi-infinite
2. Both the palm and the object have constant material properties
3. Heat transfer is one-dimensional
4. 108
Neglecting thermal contact resistance
The 1-dimensional Heat Equation can be written as follows:
\begin{equation} \frac{\partial^2T}{\partial x^2} = \frac{1}{\alpha}\frac{\partial T}{\partial t} \end{equation}
where $\alpha = k/(\rho c)$ is the thermal diffusivity, $k$ is the thermal conductivity, $\rho$ is the density, and $c$ is the specific heat capacity. This equation describes the movement of heat within your hand where $x = 0$ is the contact point. The solution to this equation gives us the temperature $T$ as a function of $x$. The only boundary condition we will impose is that of the constant temperature wall $T(x=0)=T_w$.
Because of the semi-infinite approximation, we can use some tricks to solve this equation and get a very nice result:
\begin{equation} \frac{T(x) - T_w}{T_i - T_w} = \mathrm{erf} \left( \frac{x}{2\sqrt{\alpha t}} \right) \end{equation}
where $\mathrm{erf}$ is the error function.
Definition of Thermal Effusivity
Now it turns out that when two semi-infinite surfaces come into contact, they must reach a constant temperature at the contact point, $T_s$ so the above solution holds for this case as well (replace $T_w$ with $T_s$). Furthermore, the energy leaving the warmer solid must equal the energy entering the cooler one. Invoking <PERSON>'s law, we get:
\begin{equation} \frac{-k_A \left( T_s - T_{A,i} \right)}{\sqrt{\pi \alpha_A t}} = \frac{-k_B \left( T_s - T_{B,i} \right)}{\sqrt{\pi \alpha_B t}} \end{equation}
where the subscripts $A$ and $B$ denote thermal properties and initial temperatures of the two solids $A$ and $B$.
If we re-arrange this equation to solve for the surface temperature, we get:
\begin{equation} T_s = \frac{\sqrt{k_A \rho_A c_A}T_{A,i} + \sqrt{k_B \rho_B c_B}T_{B,i}}{\sqrt{k_A \rho_A c_A} + \sqrt{k_B \rho_B c_B}} \end{equation}
where we define $\varepsilon \equiv \sqrt{k \rho c}$. This new property is called the thermal effusivity. It behaves as a weighting factor which 'competes' with the effusivity of the other object. 749