A metal rod AB of length 10x has its one end A in ice at 0°C and the other end B in water at 100°C.

Q: Why does heat flow from P to both A and B?

Heat always flows from a region of higher temperature to a region of lower temperature. Since P is at $400^\circ\text{C}$, which is higher than both $0^\circ\text{C}$ (at A) and $100^\circ\text{C}$ (at B), heat will naturally conduct outwards from P towards both ends.

Q: Why do we use $10x - \lambda x$ for the length of PB?

The total length of the rod is given as $10x$. If the distance from A to P is $\lambda x$, then the remaining length from P to B must be the total length minus $\lambda x$, which is $10x - \lambda x = (10 - \lambda)x$.

Q: What happens to the constants $k$, $A$, and $x$?

Because the rod is uniform, the thermal conductivity $k$ and cross-sectional area $A$ are the same for both segments. The variable $x$ is just a scaling factor for length. When we equate the mass rates, these common factors appear on both sides of the equation and neatly cancel out!

Q: Why is the latent heat of vaporization so much higher than the latent heat of fusion?

Melting ice (fusion) only requires breaking some of the hydrogen bonds to turn a solid into a liquid. Evaporating water (vaporization) requires completely breaking all intermolecular bonds to separate the molecules into a gas, which takes significantly more energy.

Q: Does the position of P make physical sense?

Absolutely! Since evaporating water takes much more energy ($540\text{ cal/g}$) than melting ice ($80\text{ cal/g}$), we need a much higher rate of heat flow towards B to achieve the same mass rate. A shorter length for PB means lower thermal resistance, allowing this massive heat flow. Hence, P is very close to B ($\lambda = 9$).

Question

A metal rod $AB$ of length $10x$ has its one end $A$ in ice at $0^\circ	ext{C}$ and the other end $B$ in water at $100^\circ	ext{C}$. If a point $P$ on the rod is maintained at $400^\circ	ext{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540	ext{ cal g}^{-1}$ and latent heat of melting of ice is $80	ext{ cal g}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. (Neglect any heat loss to the surrounding.)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a uniform metal rod $AB$ with a total length of $10x$. The environment at end $A$ is a block of ice at $0^\circ	ext{C}$, while end $B$ is submerged in boiling water at $100^\circ	ext{C}$.

We introduce a heat source at a specific point $P$ on the rod, maintaining it at a scorching $400^\circ	ext{C}$. Because heat naturally flows from a region of higher temperature to a region of lower temperature, thermal energy will conduct outwards from point $P$ towards both ends $A$ and $B$.

Let the distance from end $A$ to point $P$ be $\lambda x$. Consequently, the remaining distance from point $P$ to end $B$ must be $(10 - \lambda)x$.

### The Heat Conduction Equations

The rate of heat conduction through a material is given by Fourier's Law:

$$ \frac{dQ}{dt} = \frac{kA\Delta T}{L} $$

Let's analyze the heat flowing towards end $A$. This thermal energy is responsible for melting the ice. The rate of heat flow $\frac{dQ_1}{dt}$ is:

$$ \frac{dQ_1}{dt} = \frac{k A (400 - 0)}{\lambda x} $$

This heat rate provides the energy for the phase change of ice, which can be expressed using the latent heat of fusion ($L_{	ext{ice}} = 80	ext{ cal/g}$):

$$ \frac{dQ_1}{dt} = L_{	ext{ice}} \frac{dm_{	ext{ice}}}{dt} $$

Similarly, the heat flowing towards end $B$ evaporates the water. The rate of heat flow $\frac{dQ_2}{dt}$ is:

$$ \frac{dQ_2}{dt} = \frac{k A (400 - 100)}{(10 - \lambda)x} = \frac{k A (300)}{(10 - \lambda)x} $$

This heat rate provides the energy for the phase change of water, using the latent heat of vaporization ($L_{	ext{water}} = 540	ext{ cal/g}$):

$$ \frac{dQ_2}{dt} = L_{	ext{water}} \frac{dm_{	ext{water}}}{dt} $$

### Equating the Mass Rates

The problem provides a beautiful constraint: the mass of ice melting per unit time is exactly equal to the mass of water evaporating per unit time.

$$ \frac{dm_{	ext{ice}}}{dt} = \frac{dm_{	ext{water}}}{dt} $$

By substituting our heat flow equations into this constraint, we get:

$$ \frac{1}{L_{	ext{ice}}} \frac{dQ_1}{dt} = \frac{1}{L_{	ext{water}}} \frac{dQ_2}{dt} $$

$$ \frac{1}{80} \left( \frac{k A \cdot 400}{\lambda x} ight) = \frac{1}{540} \left( \frac{k A \cdot 300}{(10 - \lambda)x} ight) $$

### The Final Calculation

Notice how the physical constants of the rod—thermal conductivity $k$, cross-sectional area $A$, and the length scale $x$—appear on both sides of the equation. They neatly cancel out, leaving us with a pure algebraic relationship for $\lambda$:

$$ \frac{400}{80 \lambda} = \frac{300}{540 (10 - \lambda)} $$

Simplifying the fractions:

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

Cross-multiplying yields:

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

$$ 90 - 9\lambda = \lambda $$

$$ 10\lambda = 90 \implies \lambda = 9 $$

**Conclusion:** The point $P$ must be located at a distance of $9x$ from the ice end $A$. This makes perfect physical sense! Because evaporating water requires significantly more energy ($540	ext{ cal/g}$) than melting ice ($80	ext{ cal/g}$), the thermal resistance to the water end must be much lower to allow a massive heat flow. Placing $P$ very close to $B$ achieves exactly this.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Heat Flow Towards End A

Heat Flow Towards End B

Equating the Mass Rates

Solving for $λ$

Final Conclusion

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Heat Conduction Equations

Equating the Mass Rates

The Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Heat Flow Towards End A

Heat Flow Towards End B

Equating the Mass Rates

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Final Conclusion

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Heat Conduction Equations

Equating the Mass Rates

The Final Calculation

Similar Questions

Solving for $λ$