Problem 85 - Master Resource Book in JEE Main Physics

Question

A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of \(0.1 \mathrm{~g}\) of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is \(1 / 4\) that of first, the rate at which ice melts in \(\mathrm{g} / \mathrm{s}\) will be (a) \(3.2\) (b) \(1.6\) (c) \(0.2\) (d) \(0.1\)

Step-by-Step Solution

Answer

The rate of ice melting is 0.2 g/s, which is option (c).

Step 1: Understand the Given Problem

We have a cylindrical rod causing ice to melt at a certain rate. When this rod is replaced by another rod with different physical properties, we need to determine the new melting rate. Specifically, the second rod is half the length, double the radius, and has one-fourth the thermal conductivity of the first rod.

Step 2: Write the Formula for Rate of Heat Transfer

The rate of heat transfer through a cylindrical rod is given by Fourier's law: \[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \]Where:- \( Q \) is the heat transfer per unit time- \( k \) is the thermal conductivity- \( A \) is the cross-sectional area- \( L \) is the length of the rod- \( T_1 \) and \( T_2 \) are the temperatures at both ends.

Step 3: Calculate the Area of the Rods

For a rod of radius \( r \), the cross-sectional area \( A \) is \( \pi r^2 \). - The first rod has area \( A_1 = \pi r^2 \).- The second rod has double the radius, so \( A_2 = \pi (2r)^2 = 4\pi r^2 \).

Step 4: Calculate the Heat Transfer for Each Rod

The initial rate of heat transfer for the first rod is given as it melts \(0.1 \text{ g/s}\). Thus,\[ Q_1 = \frac{k \cdot \pi r^2 \cdot (T_1 - T_2)}{L} = 0.1 \text{ g/s} \times L_f \]Where \( L_f \) is the latent heat of fusion of ice.For the second rod,\[ Q_2 = \frac{\frac{k}{4} \cdot 4\pi r^2 \cdot (T_1 - T_2)}{\frac{L}{2}} \]Simplify to \[ Q_2 = \frac{k \cdot \pi r^2 \cdot (T_1 - T_2)}{\frac{L}{2}} \]

Step 5: Determine the Ratio of Heat Transfer

Find the ratio \( \frac{Q_2}{Q_1} \) using the simplified equations:\[ \frac{Q_2}{Q_1} = \frac{\frac{2k \cdot \pi r^2 \cdot (T_1 - T_2)}{L}}{\frac{k \cdot \pi r^2 \cdot (T_1 - T_2)}{L}} = 2 \]This shows the heat transfer with the second rod is twice as much as with the first.

Step 6: Calculate New Melting Rate

Since the rate of heat transfer is twice for the second rod, the rate of ice melting will also be twice.The initial rate was \(0.1 \text{ g/s}\), so the new rate is:\[ Q_2 / L_f = 0.2 \text{ g/s} \]

Step 7: Compare with Options

The calculated melting rate is \(0.2 \text{ g/s}\), which matches option (c). This confirms that our calculation is correct.

Key Concepts

Heat Transfer

Heat transfer is a fundamental process where thermal energy is exchanged between physical systems. In the context of this exercise, the cylindrical rod functions as a medium for heat transfer from the steam chamber to the ice. There are three primary modes of heat transfer: conduction, convection, and radiation. In this scenario, conduction is primarily at play.

• Conduction: Heat is transferred through a material with no movement of the material itself. In our case, the cylindrical rod conducts heat from the steam to the ice.
• Convection: Usually occurs when a fluid, such as air or water, moves, carrying heat with it. This is not relevant for the solid rod directly.
• Radiation: The transfer of energy via electromagnetic waves. This would be a minor influence compared to conduction in this setup.

Understanding how heat conducts through the rod is essential to determining how much ice it can melt.

Fourier's Law

Fourier's Law provides us with a mathematical approach to predict the rate at which heat is conducted through a material. It is expressed as: \[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \]

• Thermal Conductivity \( k \): This represents how well a material can conduct heat. A high value means the material is a good conductor.
• Cross-sectional Area \( A \): For cylindrical rods, this is calculated using \( \pi r^2 \), where \( r \) is the radius of the rod.
• Temperature Difference \( (T_1 - T_2) \): The difference in temperature between the steam chamber and the ice significantly affects the rate of heat transfer.
• Length \( L \): The length of the rod affects how quickly heat travels from one end to the other.

In this exercise, we observe how altering these parameters—by changing the rod dimensions—affects heat flow and consequently the melting rate of the ice.

Latent Heat of Fusion

The latent heat of fusion is an important thermodynamic concept. It is the amount of heat required to change a substance from solid to liquid without a temperature change. In this scenario, the latent heat of fusion of ice is crucial as it dictates how much heat must be transferred through the rod to melt a specified quantity of ice.
The formula that connects heat transfer rate to the melting rate is: \[ Q = m \times L_f \] Where:

• \( Q \) is the heat transfer rate,
• \( m \) is the mass of the ice melted per second,
• \( L_f \) is the latent heat of fusion of ice.

In our problem, the rate of heat transfer \( Q \) determines the melting rate \( m \), utilizing the latent heat of fusion to convert this energy into the phase change from solid to liquid.

Cylindrical Rod

The cylindrical rod serves as a conduit for heat transfer between the steam chamber and the ice, utilizing its unique geometric properties.
With a cylinder, factors like the radius and the length are vital in determining the efficiency of heat transfer.

• Radius: The cross-sectional area \( A = \pi r^2 \) of the rod increases with the square of the radius, significantly affecting the heat conduction capacity.
• Length: Reducing the rod's length allows for quicker heat transfer from one end to the other since the heat travels a shorter distance.

In this exercise, the second rod has a doubled radius and half the length of the first, meaning its attributes profoundly alter the rate of heat transfer. These changes exemplify the principles of conductive heat transfer through a cylindrical medium.