Problem 116 - Master Resource Book in JEE Main Physics

Question

A brass boiler has a base area \(0.15 \mathrm{~m}^{2}\) and thickness \(1.0 \mathrm{~cm}\). It boils water at the rate of \(6.0 \mathrm{~kg} / \mathrm{min}\) when placed on a gas stove. The temperature of the part of the flame in contact with the boiler will be. (Thermal conductivity of brass \(=109 \mathrm{~J} / \mathrm{s}-\mathrm{m}-\mathrm{K}\), Heat of vapourization of water \(=2256 \times 10^{3} \mathrm{~J} / \mathrm{kg}\) ) [NCERT] (a) \(158^{\circ} \mathrm{C}\) (b) \(208^{\circ} \mathrm{C}\) (c) \(238^{\circ} \mathrm{C}\) (d) \(264^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The temperature of the flame is approximately \(238^{\circ} \mathrm{C}\).

Step 1: Understand the Problem

We need to determine the temperature of the flame in contact with a brass boiler. We know the heat of vaporization of water and the thermal conductivity of brass, as well as the boiling rate of water.

Step 2: Calculate Heat Required to Boil Water

The rate at which heat must be supplied to boil water is given by the product of the heat of vaporization and the mass of water boiled per second. The heat required is calculated as follows:\[Q = m \times L = 6.0 \, \text{kg/min} \times \frac{1}{60} \, \text{min/s} \times 2256 \times 10^{3} \, \text{J/kg} = 226,560 \, \text{J/s}\]

Step 3: Apply Fourier’s Law of Heat Conduction

Using Fourier's law, the rate at which heat is conducted through the material is:\[Q = K \times A \times \frac{\Delta T}{d}\]Where \(K\) is the thermal conductivity, \(A\) is the area, \(\Delta T\) is the temperature difference, and \(d\) is the thickness.

Step 4: Set up the Equation for Temperature Difference

Rearrange Fourier's law to solve for the temperature difference:\[\Delta T = \frac{Q \times d}{K \times A}\]Substitute the known values:\[\Delta T = \frac{226,560 \, \text{J/s} \times 0.01 \, \text{m}}{109 \, \text{J/s-m-K} \times 0.15 \, \text{m}^2}\]

Step 5: Calculate \(\Delta T\)

Calculate the temperature difference \(\Delta T\):\[\Delta T = \frac{2,265.6}{16.35} \, \text{K} \approx 138.5 \, \text{K}\]

Step 6: Determine the Flame Temperature

The boiling point of water is \(100^{\circ} \mathrm{C}\). Add \(\Delta T\) to the boiling point to find the flame temperature:\[\text{Flame temperature} = 100^{\circ} \mathrm{C} + 138.5 \, \mathrm{K} \approx 238^{\circ} \mathrm{C}\]

Step 7: Select the Correct Option

Compare the calculated flame temperature with the choices provided in the problem. The correct choice is \(238^{\circ} \mathrm{C}\) which corresponds to option (c).

Key Concepts

Fourier’s Law of Heat Conduction

In understanding the concept of how heat is transferred through materials, Fourier’s Law of Heat Conduction plays a crucial role. Essentially, this law explains how heat flows from a region of high temperature to a region of low temperature in a steady manner. The rate of heat conduction is directly proportional to the temperature gradient, area through which heat flows, and the material's thermal conductivity. In mathematical terms, Fourier's Law can be expressed as:\[ Q = K \times A \times \frac{\Delta T}{d} \]Where:- \( Q \) is the heat transferred per unit time.- \( K \) is the thermal conductivity of the material, which indicates how well a material can conduct heat.- \( A \) is the surface area through which heat is being transferred.- \( \Delta T \) is the difference in temperature across which the heat is flowing.- \( d \) is the thickness of the material through which heat is being conducted.This law was used in the problem to determine the rate at which heat is conducted through the brass boiler. It highlights the relationship between these variables and how changing one can impact the overall heat transfer.

Heat of Vaporization

The heat of vaporization is a critical concept in understanding phase changes, particularly from liquid to gas. When a substance changes state, such as water boiling and turning into steam, it absorbs heat energy without changing temperature. This absorbed energy, called the heat of vaporization, is crucial for overcoming intermolecular forces during the phase transition.For water, the heat of vaporization is particularly high at approximately \( 2256 \times 10^3 \text{ J/kg} \). This means it requires a significant amount of energy to turn liquid water into vapor. This energy is provided by a heat source, such as a flame in our problem scenario.In the calculation, the mass flow of water vaporized per minute was multiplied by this constant to determine how much heat energy is needed per second to keep the water boiling. This value is then used in subsequent calculations to understand the entire heat transfer process through the boiler.

Temperature Difference

Temperature difference is a simple yet foundational concept in thermal dynamics. It represents the driving force of heat transfer between two points. The greater the temperature difference, the more heat energy will flow per unit time.In the problem, the temperature difference \( \Delta T \) was a crucial variable in determining the temperature at the flame-boiler interface. By rearranging Fourier's Law, we calculated \( \Delta T \) as follows:\[ \Delta T = \frac{Q \times d}{K \times A} \]Using our known values: the rate of heat energy \( Q \), the thickness of the boiler \( d \), the thermal conductivity \( K \), and the area \( A \), we could find this temperature difference, which was approximately \( 138.5 \text{ K} \).With the boiling point of water at \( 100^{\circ} \text{C} \), adding the calculated temperature difference gives us the temperature of the flame that keeps the water continually boiling. Understanding this relationship emphasizes the role temperature difference plays in practical thermal applications.