Problem 91 - Master Resource Book in JEE Main Physics

Question

A metal \(\operatorname{rod} A B\) of length \(10 x\) has its one end \(A\) in ice at \(0^{\circ} \mathrm{C}\) and the other end \(B\) in water at \(100^{\circ} \mathrm{C}\). If a point \(P\) on the rod is maintained at \(400^{\circ} \mathrm{C}\), then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporationof water is \(540 \mathrm{cal} / \mathrm{g}\) and latent heat of melting of ice is \(80 \mathrm{cal} / \mathrm{g}\). 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 surroundings). (a) 9 (b) 2 (c) 6 (d) 1

Step-by-Step Solution

Answer

\(\lambda = 9\)

Step 1: Understand the problem setup

We have a rod of length \(10x\) with one end in ice at \(0^{\circ} \mathrm{C}\) and the other in water at \(100^{\circ} \mathrm{C}\). A point \(P\) on the rod is maintained at \(400^{\circ} \mathrm{C}\), and we need to find where this point \(P\) is located relative to point \(A\), as a fraction of \(x\), denoted by \(\lambda x\). Equal amounts of ice melting and water evaporating implies equal heat flows.

Step 2: Relate heat flow to latent heats

Consider the heat flow caused by the temperature at \(P\). The rate of ice melting depends on the heat flowing from \(P\) towards \(A\), while the rate of water evaporation depends on the heat flowing from \(P\) towards \(B\). Since both are equal, we have:\[ q_1 = q_2 \]Where \(q_1\) is the heat causing ice to melt and \(q_2\) is the heat causing water to evaporate.

Step 3: Set up the equality of heat flows

Heat flow for melting ice is given by:\[ q_1 = mL_f \] where \(L_f = 80 \, \mathrm{cal/g}\). Heat flow for evaporating water is given by:\[ q_2 = mL_v \] where \(L_v = 540 \, \mathrm{cal/g}\). Given \( m L_f = m L_v \), we can equate them and simplify to relate the distances to \(P\).

Step 4: Translate equality into a distance relationship

Since equal heat flows melt and evaporate equal masses of ice and water, and using the respective heat relations:\[ m \cdot 80 = m \cdot 540 \]This implies:\[ Heat \ proportionality \frac{distance \ from \ A}{distance \ from \ B} = \frac{L_v}{L_f} = \frac{540}{80} = \frac{27}{4} \].

Step 5: Solve for \( \lambda \)

Using the proportionalities established:If \(P\) is at distance \(\lambda x\) from \(A\), then it is \(10x - \lambda x\) from \(B\). Setting \[ \frac{\lambda x}{10x - \lambda x} = \frac{27}{4} \]Simplifying gives:\[ 4\lambda x = 27(10x - \lambda x) \]\[ 4\lambda = 270 - 27\lambda \]\[ 31\lambda = 270 \]\[ \lambda = \frac{270}{31} \]

Step 6: Final calculation and checking options

The calculated \(\lambda\) approximately equals \(8.71\), which rounds to \(9\) as the closest integer matching the given options. Thus, option (a) is correct, \(\lambda = 9\).

Key Concepts

Latent Heat

Latent heat is a fascinating concept in thermodynamics. It refers to the heat absorbed or released by a substance during a phase change without changing its temperature. Understanding this helps clarify why, even at constant temperature, energy is needed to melt ice or evaporate water.

• Latent Heat of Fusion: This is the heat required to change a unit mass of a solid into a liquid at its melting point. In our problem, it amounts to 80 cal/g for ice, indicating the energy needed to melt one gram of ice without rising temperature.
• Latent Heat of Vaporization: This is the energy required to convert a unit mass of a liquid into vapor at its boiling point. For water, this is 540 cal/g, signifying the heat required to evaporate one gram of water.

When working with problems involving latent heat, remember that even if temperatures remain steady, energy is still in motion, reshaping the state of the material.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one object or substance to another. In our rod scenario, understanding how heat moves along the metal is crucial. Thermal conductivity measures how well a material facilitates this process.
There are three methods of heat transfer: conduction, convection, and radiation, but since our problem involves a metal rod, we focus on conduction. Conduction occurs when molecules in a hot area of a material collide with neighboring, cooler molecules, transferring energy.
In terms of physics, the law governing conduction can be formulated as:

• Heat Flow = Thermal Conductivity × Area × Temperature Difference / Distance

In our exercise, the metal rod facilitates heat flow, causing ice to melt on one end and water to evaporate on the other. The equality of these heat flows at point P is essential for solving the problem.

Phase Change

A phase change happens when a substance transitions between solid, liquid, or gaseous states. These transformations are central to many natural processes and are energetically significant because they require or release latent heat.
In the problem at hand, the phase change of both ice melting into water and water turning into vapor are occurring. This is key because these actions explain why energy transitions are happening without temperature change. Recognizing the specific amounts of energy required for each phase change is necessary for solving related physics problems.

• Melting: Ice changing to water requires input energy. Despite the same temperature, energy intake causes rearrangement of molecules into a liquid form.
• Evaporation: Water converting to vapor also needs additional energy without a shift in temperature. This accounts for the energy transferred through conduction from the rod.

Understanding these changes helps clarify how the rod's maintained temperature at point P causes equal rates of energy-based transformations at the respective ends.