Physics, 16.11.2020 17:00 yeet1273

One object has a temperature twice as large as another object. If the objects have the same surface area, how much more power does the hotter object radiate than the cooler object

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Answer from: scholar13

The hotter object radiates 15 times more power than the power of cooler object.

Absolute Temperature of one object = T_1

Absolute Temperature of second object = T_2 = 2T_1

The Power emitted by the an object is given by the equation (1)

$P= A\epsilon \sigma\;T^4$ .......(1)

Equation (1) is called as Stephan Boltzmann Law

Where

P = Power emitted by the object in Joule

A = Surface area of the object

$\epsilon$ = Emissivity of the object

T = Absolute Temperature

Let us consider emissivities are equal

$So \; P_1/P_2 = T_1^4/T_2^4\\$ ( Areas of both objects are equal)

P_1/P_2= T_1^4/16T1^4= 1/16

P_2= 16 P_1

Difference in Power = 16P_1- P_1 = 15P_1

So the hotter object radiate 15 times more power than the power of cooler object.

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Answer from: kneari

The hotter object radiate more power than the cooler object 15 times i.e 15σeA $T^{4}$

Explanation:

From Stefan's law, an object would radiate power with respect to its temperature.

i.e Radiative power, Q = σeA $T^{4}$

where Q is the radiative power, σ is the constant, e is the emissivity of the object, A is the area of the object and T is the temperature.

Let the temperature of the cooler object be represented by T.

So that its radiative power = σeA $T^{4}$

Given that the temperature of the hotter object is twice as large as that of the cooler object.

Temperature of hotter object = 2T

So that its radiative power = σeA $(2T)^{4}$

= 16σeA $T^{4}$

Radiative power difference between the two objects = 16σeA $T^{4}$ - σeA $T^{4}$

= 15σeA $T^{4}$

The hotter object radiate more power than the cooler object 15 times.

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