Q. 54.0( 2 Votes )
Two bodies A and B having equal surface areas are maintained at temperatures 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio
A. 1: 1.15
B. 1: 2
C. 1: 4
D. 1: 16
The thermal radiation emitted in a given time by A and B are in the ratio 1: 1.15.
Stefan-Boltzmann Law is given as:
The energy of thermal radiation emitted per unit time by a body having surface area A is given as:
Here, e (between 0 to 1) is the emissivity of the body and σ is the Stefan-Boltzmann constant and T is the Temperature.
Now the temperature should be in Kelvin.
For body A :
For body B :
Here TA = 273 + 10 ° C = 283 K.
TB = 273 + 20° C= 293 K
Substituting we get,
Out of all the options only 1:1.15 gives answer close to 0.8703 hence,
Thus option (a) is the correct option.
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