Question

Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount by which the star's thermal radiation increases the entropy of the entire universe each second. Assume that the star is a perfect blackbody, and that the average temperature of the rest of the universe is 2.73 K. Do not consider the thermal radiation absorbed by the star from the rest of the universe. J/K

Answer 1

Answer:

The value is  $\Delta s = 8.537 *10^{25 } \ J/K$

Explanation:

From the we are told that

   The radius of the sphere is $r = 6.32 *10^{8} \ m$

   The temperature is $T_x = 5350 \ K$

    The average temperature of the rest of the universe is  $T_r = 2.73 \ K$

Generally the change in entropy of the entire universe per second is mathematically represented as

         $\Delta s = s_r - s_x$

Here $s_r$ is the entropy of the rest of the universe which is mathematically represented as

          $s_r = \frac{Q}{T_r}$

Here Q is the quantity of heat radiated by the star which is mathematically represented as

           $Q = 4 \pi * r^2 * \sigma * T^4_x$

Here $\sigma$ is the Stefan-Boltzmann constant with value  

           $\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.$

=>         $Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4$

=>         $Q = 2.332 *10^{26} \ J$

So

      $s_r = \frac{2.332 *10^{26}}{2.73}$

=>   $s_r = 8.5415 *10^{25}\ J/K$

Here $s_x$ is the entropy of the rest of the universe which is mathematically represented as

      $s_x = \frac{Q}{T_x}$

=>   $s_x = \frac{2.332 *10^{26} }{5350}$

=>   $s_x = 4.359 *10^{22} \ J/K$

So

      $\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}$

=>   $\Delta s = 8.537 *10^{25 } \ J/K$

Answer 2

This question involves the concepts of entropy and the thermal radiation

The entropy of the entire universe is increased by "8.41 x 10²⁵ J/k

".

The increase in entropy is given as follows:

$\Delta s = s-s_T$

where,

Δs = increase in entropy = ?

σ = Stefan-Boltzman's constant = 5.67 x 10⁻⁸ W/m².k⁴

A = surface area = 4πr² = 4π(6.32 x 10⁸ m)² = 5.01 x 10¹⁸ m²

Tr = Absolute temperature of the star = 5350 K

T = absolute temperature of the rest of the universe = 2.73 k

Q = thermal radiation energy

Q = $\sigma A T_r^4=(5.67\ x\ ^{-8}\ W/m^2.k^4)(5.01\ x\ ^{18}\ m^2)(5350\ k)^4=2.3\ x\ 10^{26}\ J$

s = entropy of the universe = $\frac{Q}{T}=\frac{2.3\ x\ 10^{26}\ J}{2.73 k}=8.42\ x\ 10^{25}\ J/k$

$s_T$ = entropy of the star = $\frac{Q}{T_r}=\frac{2.3\ x\ 10^{26}\ J}{5350\ k}=4.3\ x\ 10^{22}\ J/k$

Therefore,

Δs = 8.42 x 10²⁵ J/k - 4.3 x 10²² J/k

Δs = 8.41 x 10²⁵ J/k

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