Question

The absolute temperature of a perfectly black body is increased to twice its value. The rate of emission of energy per unit area will be

Answer 1

Explanation:

The absolute temperature of a perfectly black body is increased to twice its value. The rate of emission of energy per unit area will be

Answer 2

The correct answer is 16.

Given: Temperature is doubled.

To Find: The rate of emission of energy per unit area.

Solution:

As we know,

$P = e$ νΑ$T^{4}$  (v is gamma here)

Now, for initial case.

$P_{1} =evAT_{1} ^{4}$

Now, for final case.

$P_{2} =evAT_{2} ^{4}$

In question it is given that

$2T_{1}= T_{2}$

Therefore,

$\frac{P_{2} }{P_{1} } =(\frac{T_{2} }{T_{1} } )^{4}$

$P_{2}=16P_{1}$

Hence, the rate of emission of energy per unit area is 16 times.

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