1. Set up heat flow on both sides of point $P$

A rod $AB$ has total length $10x$.

• End $A$ is at $0^\circ C$
• End $B$ is at $100^\circ C$
• Point $P$ is maintained at $400^\circ C$

Let the distance of $P$ from $A$ be $AP=\lambda x$. Then, $PB=10x-\lambda x=(10-\lambda)x$

Since the rod is uniform, thermal conductivity $k$ and cross-sectional area $A$ are same throughout.

2. Heat conducted from $P$ to the ice end

Temperature difference across $AP$ is $400-0=400^\circ C$

So heat conducted per unit time toward ice: $\dot Q_1=\frac{kA(400)}{\lambda x}$

This heat melts ice at $0^\circ C$. If $m_1$ grams of ice melt per unit time, then $\dot Q_1=m_1 L_f=m_1\cdot 80$

Hence, $m_1=\frac{\dot Q_1}{80}$

3. Heat conducted from $P$ to the water end

Temperature difference across $PB$ is $400-100=300^\circ C$

So heat conducted per unit time toward water: $\dot Q_2=\frac{kA(300)}{(10-\lambda)x}$

This heat evaporates water at $100^\circ C$. If $m_2$ grams of water evaporate per unit time, then $\dot Q_2=m_2 L_v=m_2\cdot 540$

Hence, $m_2=\frac{\dot Q_2}{540}$

4. Use the given condition

Given that equal amounts of ice melt and water evaporate per unit time, $m_1=m_2$

Therefore, $\frac{\dot Q_1}{80}=\frac{\dot Q_2}{540}$

Substitute $\dot Q_1$ and $\dot Q_2$: $\frac{1}{80}\cdot \frac{kA(400)}{\lambda x}=\frac{1}{540}\cdot \frac{kA(300)}{(10-\lambda)x}$

Cancel common factors $kA$ and $x$: $\frac{400}{80\lambda}=\frac{300}{540(10-\lambda)}$

Simplify: $\frac{5}{\lambda}=\frac{5}{9(10-\lambda)}$

Cancel $5$: $\frac{1}{\lambda}=\frac{1}{9(10-\lambda)}$

So, $\lambda=9(10-\lambda)$

$\lambda=90-9\lambda$

$10\lambda=90$

$\lambda=9$

5. Final answer

$\boxed{9}$