Problem 72 - Master Resource Book in JEE Main Physics

Question

Two identical conducting rods are first connected independently to two vessels, one containing water at \(100^{\circ} \mathrm{C}\) and the other containing ice at \(0^{\circ} \mathrm{C}\). In the second case, the rods are joined end to end and connected to the same vessels. Let \(q_{1}\) and \(q_{2} \mathrm{~g} / \mathrm{s}\) be the rate of melting of ice in two cases respectively. The ratio of \(q_{1} / q_{2}\) is (a) \(1 / 2\) (b) \(2 / 1\) (c) \(4 / 1\) (d) \(1 / 4\)

Step-by-Step Solution

Answer

The ratio \(q_1/q_2\) is 2, corresponding to option (b) \(2/1\).

Step 1: Understanding the Heat Transfer in First Case

In the first case, both rods are connected independently to the vessels. Each rod conducts heat directly from the hot vessel at \(100^{\circ} \mathrm{C}\) to the cold vessel at \(0^{\circ} \mathrm{C}\). The steady-state heat transfer rate through each rod is given by Fourier’s law of heat conduction, which is \(Q_1 = \frac{kA(T_1 - T_2)}{L}\), where \(k\) is the thermal conductivity of the rod, \(A\) is the cross-sectional area, \(T_1 - T_2\) is the temperature difference (\(100 - 0 = 100^{\circ}C\)), and \(L\) is the length of the rod.

Step 2: Calculating Heat Transfer in Second Case

In the second case, the rods are arranged in series, so the effective length for heat conduction becomes \(2L\). The heat transfer rate for the same temperature difference \(T_1 - T_2 = 100^{\circ}C\) is now \(Q_2 = \frac{kA(T_1 - T_2)}{2L}\). Therefore, because the length doubles, the heat conduction rate halves, so \(Q_2 = \frac{1}{2}Q_1\).

Step 3: Relating Heat Transfer to Rate of Melting

The rate of melting of ice \(q\) is directly proportional to the heat transfer rate \(Q\). Therefore, in the two scenarios, the rates of melting of ice \(q_1\) and \(q_2\) relate as \(q_1 = Q_1\) and \(q_2 = Q_2 = \frac{1}{2}Q_1\).

Step 4: Finding the Ratio q1/q2

Using the relationship from Step 3, we find the ratio of \(q_1\) to \(q_2\) by \(\frac{q_1}{q_2} = \frac{Q_1}{\frac{1}{2}Q_1} = 2\). Thus, \(\frac{q_1}{q_2} = 2\).

Key Concepts

Fourier’s Law

Fourier’s Law is a fundamental principle of heat conduction. It explains how heat flows through a material. The key idea is that heat flows from a region of higher temperature to a region of lower temperature.
For a given material, the heat transfer rate (Q) is determined by three factors:

• The thermal conductivity (k), which indicates how well a material can conduct heat.
• The area (A) through which heat is being transferred.
• The temperature difference across the material ((T_1 - T_2)).

The law is expressed mathematically as:\[Q = \frac{kA(T_1 - T_2)}{L}\]Here, L is the length of the material through which heat is conducted. In the exercise, this formula helps us calculate the heat transfer when each rod connects the hot and cold vessels independently.

Thermal Conductivity

Thermal conductivity is a property that indicates a material's ability to conduct heat. It is denoted by the symbol \(k\). Materials with high thermal conductivity transfer heat more efficiently, while materials with low thermal conductivity act as insulators.
In the context of the exercise, the rods made of a conducting material have certain thermal conductivity \(k\). This value affects how much heat can be transferred through the rods when subjected to a temperature gradient.
When calculating the heat transfer rates \(Q_1\) and \(Q_2\) in the exercise, thermal conductivity is crucial as it appears in the formula:\[Q = \frac{kA(T_1 - T_2)}{L}\]By understanding a material's thermal conductivity, we can predict how effective the rods are in transferring heat between the two vessels under different setups.

Steady-State Heat Transfer

Steady-state heat transfer refers to a condition where the temperature distribution in a material does not change with time. This means the heat entering one side of a material equals the heat leaving the other side.
In the exercise, steady-state is assumed for both scenarios with the rods. This means the amount of heat transferred from the hot vessel to the cold vessel remains constant, allowing the use of Fourier’s Law to find heat transfer rates.
The concept ensures that calculations are simplified because we do not need to consider changes over time. Instead, we calculate the heat transfer rate directly based on the constant temperature difference and material properties like length and conductivity. This assumption is crucial for deriving the melting rates of ice \(q_1\) and \(q_2\) in the problem.