
==> $\frac{{dm}}{{dt}} = \frac{{KA}}{l}\,\left( {\frac{{\Delta \theta }}{L}} \right)$. Let the desire point is at a distance $x$ from water at $100°C$ .
( Rate of ice melting = Rate at which steam is being produced
==> ${\left( {\frac{{dm}}{{dt}}} \right)_{Steam}} = {\left( {\frac{{dm}}{{dt}}} \right)_{Ice}}$
==> ${\left( {\frac{{\Delta \theta }}{{Ll}}} \right)_{Steam}} = {\left( {\frac{{\Delta \theta }}{{Ll}}} \right)_{Ice}}$
==> $\frac{{(200 - 100)}}{{540 \times x}} = \frac{{(200 - 0)}}{{80\,(3.1 - x)}}$
==> $x = 0.4 m = 40 cm$
