Temperature given a ratio of power out/power in

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SUMMARY

The discussion centers on calculating the average temperature of Jupiter's cloud tops based on its energy emission and absorption. Given that Jupiter emits approximately twice the energy it receives from the Sun, the average temperature must be derived from the equation [T(observed)/T(calculated)] ^ 4 = Power out/ Power in. The calculated temperature of 107 K corresponds to the energy absorbed, while the actual temperature, factoring in the excess energy emission, is determined to be approximately 90 K. This indicates that Jupiter's internal heat processes significantly influence its thermal dynamics.

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Homework Statement


Statement: If Jupiter emitted just as much energy per second (as infared radiation) as it receives from the Sun, the average temperature of the planet’s cloud tops would be about 107 K. Given that Jupiter actually emits approximately twice this much energy per second, calculate what the average temperature must actually be.

Homework Equations


I am given an example that T(Jup.) = 103 K, the temperature of the planet as a total blackbody is T = 127 K, and therefore
[T(observed)/T(calculated)] ^ 4 = Power out/ Power in

In this specific example then, I have... [127/103]^4 = 2.3 which means that Jupiter emits 2.3 times the power than it absorbs

The Attempt at a Solution


How am supposed to go about this problem. First, I thought that if I make Tcalc = 107 K and set (Tobs./107)^4 = 1 (from the first part of the statement), then I can calculate Tobs. and use it. This would mean that To must be 107 too.

Then, if I plug (107/Tcalc)^4 = 2 (for twice as much energy/s) my Tcalc will come out to approximately 90K.

Is this all i need to do? 90K does not seem right to me. Suggestions would help me tremendously! Thanks
 
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The emitted energy per unit time is proportional to the forth power of the body. It is the same as the absorbed power in equilibrium, and this power would correspond to 107 K.
Pabs= const*1074

Jupiter is not in thermal equilibrium as there are heat productive processes inside it. Because of this processes, it emits twice as much energy as it absorbs. Its emission rate is proportional to the fourth power of the real temperature of its clouds.

Pemission =const*T4=2*Pabs.

ehild
 

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