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Question

A brass rod (thermal conductivity 109 J/s m K) has an area of cross section 0.04 m 2and length 20 cm. If the two end of the rod are maintained at a temperature difference of 200°C, the rate of heat flow through the rod is ________.

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is 4.36 kJ/s

Understanding Heat Flow Through a Brass Rod

The question asks us to calculate the rate of heat flow through a brass rod given its dimensions, thermal conductivity, and the temperature difference across its ends. This is a classic application of Fourier's Law of Heat Conduction.

Fourier's Law of Heat Conduction:

The rate of heat flow (\(\frac{Q}{t}\)) through a material is directly proportional to the area of cross-section (\(A\)), the temperature difference (\(\Delta T\)), and the thermal conductivity (\(k\)) of the material, and inversely proportional to the length (\(L\)) of the material. The formula is given by:

\(\frac{Q}{t} = k A \frac{\Delta T}{L}\)

where:

  • \(\frac{Q}{t}\) is the rate of heat flow (in J/s or Watts)
  • \(k\) is the thermal conductivity of the material (in J/s m K)
  • \(A\) is the area of the cross-section (in m<sup>2</sup>)
  • \(\Delta T\) is the temperature difference across the ends (in K or &deg;C)
  • \(L\) is the length of the rod (in m)

Given Parameters for the Brass Rod

Let's list the values provided in the question:

  • Thermal conductivity of brass, \(k = 109 \, \text{J/s m K}\)
  • Area of cross-section, \(A = 0.04 \, \text{m}^2\)
  • Length of the rod, \(L = 20 \, \text{cm}\)
  • Temperature difference, \(\Delta T = 200 \, \text{&deg;C}\)

Converting Units

Before plugging the values into the formula, we need to ensure all units are consistent. The length is given in centimeters, so we convert it to meters:

\(L = 20 \, \text{cm} = 20 \times 10^{-2} \, \text{m} = 0.20 \, \text{m}\)

The temperature difference is given in &deg;C. For a temperature difference, the value in Kelvin is the same as in Celsius. So, \(\Delta T = 200 \, \text{K}\).

Calculating the Rate of Heat Flow

Now, substitute the values into the formula for the rate of heat flow:

\(\frac{Q}{t} = k A \frac{\Delta T}{L}\)

\(\frac{Q}{t} = (109 \, \text{J/s m K}) \times (0.04 \, \text{m}^2) \times \frac{200 \, \text{K}}{0.20 \, \text{m}}\)

Let's calculate the numerical value:

\(\frac{Q}{t} = 109 \times 0.04 \times \frac{200}{0.2}\)

\(\frac{Q}{t} = 109 \times 0.04 \times 1000\)

\(\frac{Q}{t} = 109 \times 40\)

\(\frac{Q}{t} = 4360 \, \text{J/s}\)

Converting to kJ/s

The options are given in kilojoules per second (kJ/s). We convert the result from J/s to kJ/s:

\(1 \, \text{kJ} = 1000 \, \text{J}\)

So,

\(4360 \, \text{J/s} = \frac{4360}{1000} \, \text{kJ/s} = 4.360 \, \text{kJ/s}\)

The rate of heat flow through the brass rod is 4.36 kJ/s.

Comparing with Options

Let's compare our calculated rate of heat flow with the given options:

  • Option 1: 3.42 kJ/s
  • Option 2: 2.32 kJ/s
  • Option 3: 4.36 kJ/s
  • Option 4: 5.80 kJ/s

Our calculated value of 4.36 kJ/s matches Option 3.


Revision Table: Heat Conduction Concepts

Concept Description Formula
Heat Conduction Transfer of heat through direct contact, occurring in solids, liquids, and gases (most significant in solids). N/A
Thermal Conductivity (\(k\)) A material's ability to conduct heat. Higher \(k\) means better conductor. Units: J/s m K or W/m K
Rate of Heat Flow (\(\frac{Q}{t}\)) The amount of heat energy transferred per unit time. \(\frac{Q}{t} = k A \frac{\Delta T}{L}\) (for steady-state conduction)
Temperature Gradient (\(\frac{\Delta T}{L}\)) The change in temperature per unit length along the direction of heat flow. Units: K/m or &deg;C/m

Additional Information on Heat Transfer

Heat transfer is the movement of thermal energy from a region of higher temperature to a region of lower temperature. There are three primary modes of heat transfer:

  1. Conduction: Heat transfer through direct contact between particles. This is the main mode of heat transfer in solids. Energy is transferred as vibrating atoms and molecules collide with their neighbors. Metals are generally good conductors because of free electrons.
  2. Convection: Heat transfer through the movement of fluids (liquids or gases). As a fluid is heated, it becomes less dense and rises, while cooler, denser fluid sinks, creating convection currents. This is how heat is transferred in boiling water or through air circulation.
  3. Radiation: Heat transfer through electromagnetic waves. This mode does not require a medium and can occur through a vacuum, such as heat from the sun reaching the Earth. All objects above absolute zero temperature emit thermal radiation.

In the case of the brass rod, the heat transfer is primarily through conduction because it is a solid material with a temperature difference across its ends.

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