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madreJ [45]
2 years ago
6

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
Physics
1 answer:
Mariulka [41]2 years ago
7 0

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

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