Question

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

Answer 1

Answer:

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.

Answer 2

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.

For more information please refer to the link below

https://brainly.com/question/23188212