Conduction, Temperature Gradient and Thermal Conductivity

Introduction

Heat transfers from one point to another due to the temperature difference between them. The three mechanisms by which heat can be transferred are conduction convection and radiation.

Conduction

When heat is transferred through the medium but without the actual motion of transfer of particles in the medium, then this process is called conduction.

When we heat the solid, then the heated particles vibrate their mean position and their kinetic energy will be increased. So, they transfer their kinetic energy to the neighboring particles and heat energy can be transferred from one point to another place through solid.

Application of Conduction

1. In winter, iron chairs appear to be colder than the wooden chairs due to conduction.
2. Due to conduction ice is packed in sawdust
3. Woolen clothes are warm because they have fine pores filled with air which prevents the heat conduction from the body to the surrounding.
4. Eskimos make a double-walled house of the blocks of ice in order to prevent transmission of heat from the house too cold surroundings.

Convection

When heat is transferred to the medium by the actual motion if the transfer of particles in the medium then this process is called convection. Convection is possible on liquid and gas.

Application of convection

1. Ventilation
   Warm airs in the rooms move upwards and moves outside from the ventilators.
2. Tradewinds
   The convection current of air blows from north-east towards the equator which is called trade wind.

Radiation

When heat is transferred without any material medium then this process is called radiation. We can get the heat energy from the solar system by this process.

Difference between Conduction, Convection, and Radiation

Temperature Gradient

Suppose a metal rod AB of length L whose ends are in thermal contact with a hot reservoir at temperature \(\theta _1\) and the cold reservoir at temperature \(\theta _2\). The sides of the rod are insulated to prevent a transfer of heat outside AB. When the steady state is reached, the temperature is measured along the length and the graph is plotted.

The rate of fall of temperature with distance in the direction of heat flow is called the temperature gradient. The temperature gradient of the rod is uniform.

$$\therefore \text {Temperature gradient} = \frac {\theta _1 - \theta _2}{L}$$

Between any two points Cand D, at a distance x,

$$ \text {Temperature gradient} = \frac {\theta _C - \theta _D}{L}$$

If the sides of the rod are not insulated, heat flows from these sides also to surroundings and heat flowing per second from C and D is smaller than that at the end A. In such case, the graph between temperature and length is a curve as in figure and the temperature gradient along the bar is not constant.

Thermal Conductivity

Let us consider a transfer of heat takes place through a body having two parallel faces separated by the certain distance 'd' having an area if cross-section of space 'A'. If \(\theta _1' \text {and} \theta_2' \) are the temperature of two faces such that\(\theta _1' >\theta_2' \). Then the quantity of heat 'Q' transferred from one face to another face depends upon the following factors:

1. Heat transfer is directly proportional to the difference in temperature of two faces . i.e.
   $$Q \propto (\theta_1 - \theta_2) \dots (i)$$
2. Heat transfer is directly proportional to the area if cross-section of each face i.e.
   $$Q \propto A \dots (ii)$$
3. Heat transfer is directly proportional to the time taken to transfer the heat from one face to another face. i.e
   $$Q \propto t \dots(iii) $$
4. Heat transfer is directly proportional to the distance between two faces. i.e.
   $$Q \propto \frac {1}{d} \dots (iv)$$

By combining equation (i), (ii), (iii) and (iv)

$$Q \propto A(\theta_1 - \theta_2) \frac {t}{d}$$

$$\therefore Q = \frac {KA(\theta_1 - \theta_2) t}{d}$$

where K is constant and is known as thermal conductivity of the substance. It's value depends upon the nature of a material.

The rate of transfer of heat between two faces is given by

$$\frac {Q}{t} = \frac {KA(\theta_1 - \theta_2)}{d}\dots (v)$$

From equation(v)

$$ \text {If} A = 1m^2, d = 2 m, \theta_1 - \theta_2 = 1 K \text {and} t = 1 sec $$

Then, \(\theta = K\)

Hence, the thermal conductivity of the substance can be defined as the quantity of heat transferred through the surface having 1 m² area of cross-section of each, separated by a distance of 1 m and 1 K difference in temperature between them in 1 second time.