Thermal Conductivity by Searle's Method and Heat Radiation

Determine the thermal Conductivity of the Substance by using Searle's Method

It consists of a metallic rod having two gaps is known as the distance of separation. The rod is heated by circulating stem and next is cooled by circulating cold water. Two thermometers 'T₃' and 'T₄' are placed at inlet and outlet of water and next two thermometers'T' and 'T₁' and 'T'₂' are placed in the gaps of the rod. Mercury is placed in the gaps for the proper transfer of heat to the thermometer. The wool is surrounded by wool or fur such that the rod avoids the loss of heat.

When the temperature shown by the thermometers 'T₁', 'T₂', 'T₃' and 'T₄' becomes constant then the following observations are taken.

Here let

Distance between the holes of the metallic rod = x

Cross-sectional area of the metallic rod = A

Temperature shown by the thermometers'T₁', 'T₂', 'T₃' and 'T₄' are'θ₁', 'θ₂', 'θ₃' and 'θ₄' respectively.

Thermal conductivity of the metallic rod = K

Mass of water inlet in time 't' sec = M

Specific heat capacity of water = S_w

Then, the quantity of heat transferred to the rod in time 't' second is

$$ Q = \frac {KA(\theta_1 - \theta_2) t}{x}\dots (i)$$

$$\text {Heat gained by water} = MS_w(\theta_4 - \theta_3)\dots (ii)$$

At steady state,

$$ \text {Quantity of heat transferred} =\text {heat gained by water}$$

$$\frac {KA(\theta_1 - \theta_2) t}{x} = MS_w(\theta_4 - \theta_3)$$

$$\therefore K = \frac {MS_w(\theta_4 - \theta_3)x}{A(\theta_1 - \theta_2) t}$$

Hence, by knowing the values of 'M', 'S_w', 'x', 'A', 't','θ₁', 'θ₂', 'θ₃' and 'θ₄', thermal conductivity of metallic rod can be determined.

Inverse Square Law in Heat Radiation

Consider a heat of source 's'. Let the source of heat radiation(θ) per second in all directions.

Consider 'I₁' and 'I₂' be the intensity of heat radiations at the distance 'r₁' and'r₂' from the source of heat. Then the intensity of the heat radiations can be defined as, "The amount of heat radiation received by the surface per unit area per second." So,

$$Intensity= \frac{amount \;of\;heat\;received\;per\;second}{area}$$

$$I_1 = \frac {Q}{4\pi r_1^2 }\dots (i)$$

and $$I_2 = \frac {Q}{4\pi r_2^2}\dots (i)$$

Dividing equation (i) by (ii)

$$\frac {I_1}{I_2} = \frac{\frac {Q}{4\pi r_1^2}}{\frac {Q}{4\pi r_2^2}}$$

$$\frac {I_1}{I_2} = \frac{\frac {1}{r_1^2}}{\frac {1}{r_2^2}}$$

$$\therefore I \propto \frac {1}{r^2}$$

Hence, the intensity of heat radiation is inversely proportional to the square of the distance of the surface from the source of heat radiation which verifies the inverse square law of heat radiation.

Reflection, Transmission, and Absorption of Heat Radiation

When heat radiation falls on a surface, a part of it is reflected and a part is transmitted. Suppose Q is the heat radiation incident on a surface. As shown, R is the reflected, A is the absorbed, and T is the transmitted heat through the surface. Since total energy reflected, absorbed and transmitted per unit area per second is equal to the energy falling on the surface, then

$$R + A + T = Q$$

Dividing both sides by Q, we get

$$\frac {R}{Q} +\frac {A}{Q} +\frac {T}{Q} = 1$$

$$\text (or) r + a+ t = 1 \dots (i)$$

where,

\(r = \frac {R}{Q}\), reflectance or reflection coefficient,

\(a = \frac {A}{Q}\), absorptance or absorption coefficient

and \(t = \frac {T}{Q}\), transmittance or transmission coefficient of the surface.

Absorptance is the ratio of heat energy absorbed by the surface to the total heat energy incident on it at the same time.

Transmittance is the ratio of heat energy transmitted by the surface to the total heat energy incident on it in the same time. The reflectance r, absorptance a and transmittance t depends on the nature of the surface of the body and on the wavelength of incident radiation but not on the nature of the material of the body.