General approach to heating an object to a higher temperature than the heating element? What is the general approach to heating an object to a higher temperature than then the heating element?
For example, imagine the heating element melts at 1100C and we wish to use it to liquefy something that melts at 1500C. What is a way to do that?
 A: There are several ways to do this, but you are severely limited by the second law of thermodynamics. For example, there is no direct method of transferring heat from the heater element to the melt, as the heat loss of $\mathrm{d}Q$ from the heater gives rise to an entropy loss (shown negative) of $-\mathrm{d}Q/T_H$ (here $T_H$ is the heater temperature), whilst its absorption gives rise to an entropy gain of $+\mathrm{d}Q/T_M$ (here $T_M$ is the melt temperature); there is a nett entropy decrease in this isolated system unless $T_M < T_H$.
Thus there is probably no useful way you can do this, although the following is a theoretical possibility (I say theoretical because you would probably rather in practice work with a different, less bothersome method from the outset).
You can go forward if you have a third, low temperature heat reservoir you can expel heat to. For example, the environment about you, at an ambient temperature of $T_A\approx 300\mathrm{K}$. Suppose you extract heat $\mathrm{d}Q$ from your heater element and expel some of it $\mathrm{d}Q^\prime = (1-\eta)\,\mathrm{d}Q$ (where $\eta<1$ to the ambient, extracting work $\mathrm{d} W = \mathrm{d}Q-\mathrm{d}Q^\prime = \eta\,\mathrm{d}Q$ in the process. You then use this work to drive a laser or other electric heater which adds heat $\mathrm{d} W$ to your melt. The total system (heater, melt and environment) undergoes a nett entropy change of:
$$\mathrm{d}S=-\frac{\mathrm{d}Q}{T_H}+\frac{\mathrm{d}Q^\prime}{T_A}+\frac{\mathrm{d}W}{T_M}=\mathrm{d}Q\,\left(-\frac{1}{T_H}+\frac{1-\eta}{T_A}+\frac{\eta}{T_M}\right)\tag{1}$$
and our process, whatever it may be, is in keeping with the second law so long as $\mathrm{d}S\geq 0$, or:
$$\eta \leq \frac{(T_H-T_A)\,T_M}{(T_M-T_A)\,T_H}\tag{2}$$
So you'll have a less than totally efficient process; for $T_H = 1100^\circ{\rm C}\approx1370{\rm K}$,  $T_H = 1500^\circ{\rm C}\approx1770{\rm K}$ and $T_A = 300{\rm K}$ I make the maximum efficiency from (2) to be about $94\%$. 
Note that this is higher than if we would have simply used an ideal heat engine to extract work and used this to heat the melt indirectly; the Carnot efficiency for the above process is $1-T_A/T_H$, or about $78\%$. One way of achieving the $94\%$ would be to use an ideal heat engine to drive an ideal heat pump which transfers heat from the environment into the melt.
But you probably have a choice as to how you convert your available energy anyway, and would probably in practice choose to see a different heater element.
