- Heat Transfer
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- Heat Conduction
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- Thermal Conductivity
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- Thermal resistance and conductance
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- Convection heat transfer
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- Radiation and Black Body Radiation
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- Stefan Boltzmann law
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- Nature of thermal Radiation & Wien Displacement law
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- Kirchoff's law
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- Newton's Law of Cooling
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- The rate u
_{rad}at which an object emits energy via Electromagnetic radiation depends on objects surface area A

and temperature T in kelvin of that area and is given by

$u_{rad} = \sigma \epsilon AT^4$ ------(1)

Where

$\sigma = 5.6703 \times 10^{-8} W/m^2K^4 $

is stefan boltzmann constant and $\epsilon$ is emissivity of object's surface with value between 0 and 1.

- Black - Body radiator has emissivity of 1.0 which is an ideal limit and does not occur in nature.

- The rate $u_{abs}$ at which an object absorbs energy via thermal radiation from its environment with temperature $T_{env}$ (in kelvin) is

$u_{abs} = \sigma \epsilon AT_{env}^4$ ---(2)

Where ε is same as in equation 1

- Since an object radiate energy to the environment and absorbs energy from environment its net energy exchange due to thermal radiation is

$u=u_{rad}-u_{abs}$

$u = \sigma \epsilon A(T^4 - T_{env}^4)$ ----(3)

- u is positive if net energy is being lost via radiation. and negative if it is being absorbed via radiation

A body of emissivity (.5) and of surface area $10 \; cm^2$ is heated to 400 K and is suspended in a room at temperature 300 K. Calculate the initial rate of loss of heat from the body to the room?

Now

$u = \sigma \epsilon A(T^4 - T_{env}^4)$

Here $\epsilon=.5$

$T=400K$

$T_{env}=300K$

$A=10 \times 10^{-4} m^2$

$\sigma = 5.6703 \times 10^{-8} W/m^2K^4 $

Substituting all these values

$u=.495 W$

The surface of a body has a emissivity of .55 and area of 1.5 m

Find out the following

a. What rate of heat is radiated from the body if the temperature is 50°C

b. At what rate is radiation absorbed by the radiator when the surrounding temperature is 22°C

c What is the net rate of radiation from the body

Given σ=5.67 *10

a) Rate of radiation radiated=eσAT

b) Rate of radiation absorbed=eσAT

c)Net =Rate of radiation radiated-Rate of radiation absorbed=155 W

Class 11 Maths Class 11 Physics Class 11 Chemistry

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