Question

Absolute temperature of radiating body is halved radiating power will be reduced approximately by?​

Answer 1

Explanation:

The rate of loss of heat by radiation is given by the Stefen-boltzmann law as:

Q=σT

4

A where σ is the stefen-boltzmann constant, A is the area of the radiating body.

So Q∝T

4

.

If T is doubled, Q becomes 2

4

=16 times.

So, percentage increase in rate of heat loss by radiation is (16Q−Q)×100/Q=1500%

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Answer 2

Answer:

The power radiation will increase by a factor of 16.

Explanation:

Stefan's Law states that the heat radiated from the surface of a black body is directly proportional to the fourth power of its temperature. Where as sink temperature would remain constant if it is environment.

Consider the Stefan's Law.

$P=4 \pi r^2 \sigma T^4$

Here,

r  is the radius.

σ is the Stefan Boltzmann constant calculated as

$\sigma=5.67 X 10^-8 \ W/m^2K^4$

T  is the temperature of the body.

Given if the absolute temperature of the body is doubled. Power radiated will be.

$P'=4 \pi r^2 \sigma(2T)^4$

$P'=4\pir^2\sigma(16T^4)$

$P'=16P$

Therefore, the power radiation will increase by a factor of 16.

The rate of loss of heat by radiation is given by the Stefen-boltzmann law as:

$Q=\sigma T^4A$

where σ is the stefen-boltzmann constant, A is the area of the radiating body.

So Q∝T4 .

If T is doubled, Q becomes $2^4=16$ times.

So, percentage increase in rate of heat loss by radiation is ($(16Q-Q)* (\frac{100}{Q}) \\=(15Q)(\frac{100}{Q})\\Q=(15*100)\\Q=1500$

Reference Link

• https://brainly.in/question/15157077