Problem 117 - 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) \(264^{*} \mathrm{C}\)

Step-by-Step Solution

Answer

The temperature of the flame is \(238°C\). (Option c)

Step 1: Understand the given information

We have a brass boiler with base area \(0.15 \text{ m}^2\) and thickness \(0.01 \text{ m}\). The boiler is boiling water at a rate of \(6.0 \text{ kg/min}\). We are given the thermal conductivity of brass as \(109 \text{ J/(s-m-K)}\) and the heat of vaporization of water as \(2256 \times 10^3 \text{ J/kg}\). We need to find the temperature of the flame in contact with the boiler.

Step 2: Determining heat transfer rate

Calculate the heat needed to vaporize water at the given rate. The rate of heat transfer \(Q\) is given by:\[ Q = mL \]where \(m\) is the mass per second, and \(L\) is the heat of vaporization. First, find \(m\) by converting the rate of water to kg per second:\[ m = \frac{6.0 \text{ kg}}{60 \text{ s}} = 0.1 \text{ kg/s} \]Therefore,\[ Q = 0.1 \times 2256 \times 10^3 = 225600 \text{ J/s} \]

Step 3: Using Fourier's Law of thermal conduction

Apply Fourier's Law, which is given by:\[ Q = \frac{kA(T_2 - T_1)}{d} \]where \(Q\) is the rate of heat transfer, \(k\) is the thermal conductivity, \(A\) is the area, \(T_2\) is the flame temperature, \(T_1\) is the boiling temperature of water \(100°C\), and \(d\) is the thickness.

Step 4: Solve for the flame temperature, \(T_2\)

Substitute the known values into the Fourier's Law equation:\[ 225600 = \frac{109 \times 0.15 \times (T_2 - 100)}{0.01} \]Simplify and solve for \(T_2\):\[ 225600 = 1635(T_2 - 100) \]\[ T_2 - 100 = \frac{225600}{1635} \]\[ T_2 = \frac{225600}{1635} + 100 \]After calculating, \[ T_2 \approx 238°C \]

Step 5: Final check and answer the question

The calculated flame temperature matches with one of the given options. Thus, the temperature of the part of the flame in contact with the boiler is \(238°C\).

Key Concepts

Heat Transfer

Heat transfer is the process by which thermal energy moves from one object or substance to another. This can happen through conduction, convection, or radiation. In this exercise, we focus on conduction, which happens when heat passes through a material. The brass boiler transfers heat from the flame to the water, boiling it. The rate of heat transfer is an important factor, determining how fast water boils.

Different materials have different abilities to conduct heat. This is where thermal conductivity comes in. Brass, the material of the boiler, has a thermal conductivity of 109 J/(s-m-K), which measures how efficiently it transfers heat. The higher the thermal conductivity, the better the material at conducting heat.

• Conduction: Direct heat transfer through a material.
• Convection: Heat transfer through the movement of fluids.
• Radiation: Heat transfer through electromagnetic waves.

Understanding these methods is crucial when solving problems related to heat, like boiling water in a brass boiler.

Fourier's Law

Fourier's Law is an essential principle when it comes to understanding thermal conduction. It provides a formula to calculate the rate of heat transfer through a material. In simple terms, it says that the heat transfer rate is proportional to the temperature difference and the area through which heat flows, but inversely proportional to the material's thickness.

For the given problem, Fourier's Law is represented by the equation:
\[ Q = \frac{kA(T_2 - T_1)}{d} \]
where:

• \(Q\) is the rate of heat transfer in Joules per second (J/s).
• \(k\) is the thermal conductivity of the material, here it's brass.
• \(A\) is the area of the surface in contact with the heat.
• \(T_2\) is the temperature of the flame.
• \(T_1\) is the boiling point of water (100°C).
• \(d\) is the thickness of the material.

By using this formula, we can determine the temperature of the flame needed to achieve the boiling rate, showing the direct relationship between temperature difference and heat transfer efficiency.

Boiling Point

The boiling point is a critical concept, especially in this exercise, as it represents the temperature at which water turns into vapor. For water at standard atmospheric pressure, this occurs at 100°C. The heat required to turn water into steam is known as the heat of vaporization, and it's a significant factor when calculating the energy needed for boiling.

In the exercise we are examining, the rate at which heat is transferred into the water (225600 J/s) must meet the energy required to keep 6 kg of water boiling per minute. This makes understanding boiling points essential not just for cooking but for industrial applications as well.

The boiling process involves:

• Supplying enough heat energy to overcome intermolecular forces in the liquid.
• A constant temperature during the phase change.
• Continuing to supply heat to maintain the process of vaporization.

Understanding these components helps us grasp not only the practical sides of boiling but also provides valuable insights into thermal processes and how they can be controlled or utilized efficiently.