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3 years ago

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The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approxima

tely 293 K. What entropy change of the Universe occurs when 6.00 103 J of energy is transferred by radiation from the Sun to the Earth?

1 answer:

N76 [4]3 years ago

4 0

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, $T_s$ = 5,300 K

temperature of the Earth, $T_E$ = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

$\delta S_{sun} = \frac{-Q}{T_s}$

The earth gains Q of heat and therefore increases its entropy by the amount

$\delta S_{Earth} = \frac{-Q}{T_E}$

The total entropy change is:

$\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K$

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

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Read 2 more answers

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Answer:

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The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

$D = 4.9(3.3)^2\\ D = 53.361 m$

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