Problem 46 - 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 the material of the second rod is \(1 / 4\) that of the first, the rate at which ice melts in \(\mathrm{gs}^{-1}\) will be (a) \(3.2\) (b) \(1.6\) (c) \(0.2\) (d) \(0.1\)

Step-by-Step Solution

Answer

(c) 0.2

Step 1: Understanding the Initial Conditions

The initial rod melts 0.1 g of ice per second.

Step 2: Thermal Conductivity Equation

The rate of heat transfer through the rod is given by \[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]where \( Q \) is the rate of heat transfer, \( k \) is thermal conductivity, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, and \( L \) is the length of the rod.

Step 3: Calculating Area and Length

If the rod's radius doubles, the new cross-sectional area becomes four times the original (since area \( A = \pi r^2 \)). With half the length, the length in the equation becomes \( L/2 \).

Step 4: Impact on Thermal Conductivity

The thermal conductivity \( k \) of the second rod is \(1/4 \) of the original rod's thermal conductivity.

Step 5: Substituting Changes into Equation

Putting the changes into the equation gives:\[ Q' = \frac{(1/4) \cdot k \cdot 4A \cdot \Delta T}{L/2} = \frac{k \cdot A \cdot \Delta T}{L} \]Thus, \( Q' = 2Q \).

Step 6: Concluding the Rate of Ice Melting

Since the rate of heat transfer doubles, the rate of melting is also twice the original 0.1 g/s. Hence, the rate at which ice melts is 0.2 g/s.

Key Concepts

Rate of heat transfer

When discussing the rate of heat transfer in cylindrical rods, we are referring to how quickly heat energy moves from one end of the rod to the other. In the context of a rod with one end in a steam chamber and the other in contact with ice, this determines how fast the rod can convey heat to melt the ice. The standard equation to calculate this is given by \[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]where:

• \(Q\) is the rate of heat transfer, or the amount of heat energy transferred per unit time.
• \(k\) is the thermal conductivity of the material, indicating how easily it can transfer heat.
• \(A\) is the cross-sectional area of the rod.
• \(\Delta T\) is the temperature difference between the ends.
• \(L\) is the length of the rod.

In simple terms, this formula shows that a higher rate of heat transfer can occur with a larger cross-sectional area, a higher thermal conductivity, a greater temperature difference, or a shorter length.

Cross-sectional area of rod

The cross-sectional area of the rod plays a crucial role in determining how much heat can be transferred through it. This is because the area (\(A\) is directly proportional to the rate of heat transfer \(Q\). In cylindrical rods, the area can be calculated using the formula:\[ A = \pi r^2 \]where \(r\) is the radius of the cross-section of the rod.

Impact of Radius Change

A key observation here is how alterations to the rod's radius affect the area, and consequently the heat transfer. For instance, doubling the radius quadruples the cross-sectional area because the area is proportional to the square of the radius. This means if the radius is twice as large, there is more space for heat to pass through, significantly increasing the rate at which heat is transferred across the rod. In the exercise, changing the radius accordingly changes the thermal dynamics of the rod.

Temperature gradient across rod

The temperature gradient (\(\Delta T\)) across a rod is another vital factor influencing heat transfer. It is the difference in temperature between the two ends of the rod, which creates a driving force for the heat to flow from the hotter end (steam chamber) to the colder end (ice).

Significance of Temperature Difference

A larger temperature difference means a steeper gradient, facilitating a higher rate of heat transfer. In essence, as the temperature difference increases, so does the potential energy driving heat flow, resulting in more energy being transferred over a given time.This relationship is consistent with the laws of thermodynamics that assert that heat naturally flows from hotter to colder regions. Maintaining a high temperature gradient is essential for effective heat transfer in applications like the one described in the exercise.

Thermal properties of materials

Thermal conductivity (\(k\)) is a fundamental thermal property of materials that describes a material's ability to conduct heat. This property varies significantly across different materials and impacts how efficiently a rod can transfer heat.

Influence of Material on Heat Transfer

In the exercise, the rod's material directly influences the rate of heat transfer due to its thermal conductivity. A higher thermal conductivity implies that the material is better at transferring heat, contributing to a higher heat transfer rate.If, like in the given problem, the thermal conductivity of the second rod is a quarter (\(1/4\) ) of the original, it reduces the heat transfer capability. However, altering other factors like area and length can counterbalance this reduction, ultimately impacting the rate at which ice melts. Hence, selecting appropriate materials is crucial in designing systems that rely on efficient heat transfer.