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Viewing Single Post From: The giant space ship example

I have been thinking about this thing today and I think got an answer.
Actually I had the idea a couple days ago, but today I got "why" it was the answer.

It was when asked you how much energy it would take to heat dwarf moon to 150 K, because I
wanted to know how quickly it heated up.
And as I said it's a different way to check it. And it is. You said:
The final temperature of anything is the temperature at which it is no longer gaining thermal energy at all.

I recommend you don't worry about how long it takes to heat up and cool down. Perhaps we could make a new thread on thermodynamics in general for such questions. The question originally posed is about the power required to maintain a certain temperature in the starship. That is adding energy to make up for energy being lost as the starship radiates thermally to space.

To repeat "the final temperature of anything is the the temperature at which it is no longer gaining thermal energy at all."

Now with large thermal capacity the final temperature [if precise as within a 1/10th or say .0001 of a degree] takes a long, time.
This idea is commonly known, but maybe not appreciated enough.
Today I was asking myself the question:
If you turn off the furnaces, how long would take to cool the atmosphere.
You seem to think it would happening quickly, I don't.

To quantify this, I think it might drop in temperature fairly quickly by a couple of degrees. And by quickly I mean in a minute or two of time.
As example in solar ellipse I have been informed [never been to one] the temperature does drop rapidly. So something like 10 degrees in couple minutes. Enough that everyone notices it. Apparently.
We also have temperatures dropping by 100 K during couple hours on the Moon during a Lunar ellipse.
And we have every day a drop of about 20 C during nite every day/nite cycle on earth.
On earth we have lots of records showing daytime high and nighttime low- this amount change is not uniform. Ocean tropics has little variation. Deserts have have variation.
I would say someone who never seen earth, if they had the high and lows, humidity, and maybe barometer reading globally, could determine the output of our sun. If you gave them earth's distance from the sun, it would make it easier.
Or quickly a planet warms and cools gives you information.
Of course knowing the sun's output also gives information same type of information.
Or in case of our dwarf moon, the amount energy required to heat the moon by 1 degree over time tells you stuff. Or turning off the furnace is exact same data.
Now I don't know the formula but it will be exponential.
By which I mean, you have your 6,580,000 GigaWatts furnaces. You start will air temperature of 150 K, it will quickly heat the air- 1 K per day, but as air temperature increases it will take longer and longer to warm the air. But it's like halving the distance [sort of] it will take 1000 years to get very close to the final temperature- if what is meant by final temperature is very precise [such as .00001]. Or for practical final temperatures of say 1/10 of degree, maybe months or years.

Now, if have furnace going for months and it close to final temperature, and you turn off the furnace for 2 seconds, it may take a day or week to regain it's the lost ground.
So Earth can never get close to this kind of final temperature.
One could turn off the furnace every 12 hours and turn it back on for 12 hours and one could do this for years and reach very near it's type of final temperature. That is very similar to having furnace with half the power running constantly.

hm. That is what climate guys are doing- they treating the sun as though it's half powered furnace. Which would work if figured out how much heat was absorbed [how much the sun increased the thermal capacity- which same as it loses each nite].

So, question is how much is air temperature at surface loses per second the furnace is shut off.
If you were at final temperature the loss would be the joules produced by furnace per second minus total heat capacity.
Let's compare dwarf to earth. Dwarf 10 C. Earth desert 10 C.
Earth has higher lapse rate and per sq meter has far less air.
Night time earth desert [being 10 C]. Dwarf of course always night time.
Desert might take an hour to lower 1 C. Dwarf should take longer than 1 hr to drop 1 C.
If dwarf instead had 1/2 as much atmosphere as earth, than the temperature could
lower at faster rate than earth.

As I said above if dwarf loses some temperature from final temperature it would require a lot
of time to come back to the final temperature.
But if guess it takes 1 hr to lose 1 K, I can divide the heat capacity of atmosphere of 1 K and divide by 3600 and that how thermal power is needed. And in addition as it cools the atmosphere should drop a bit, this lowering of atmosphere should add heat. Oh but earth has more gravity and therefore would gain more energy {I thought dwarf would get more energy from this than earth}.
Anyhow, as posted above:
"So 5.4e17 KJ per K
Or 5.4e20 watts/joules And have 6,580,000 GigaWatts
Which is 6.58 x 10^15 watts
So about 100,000 seconds and one day is 86400 seconds
So 1 or two days per 1 K increase"

So if cooled at 1 K per hour, it would require 30 times more energy
than you said.
So what you are actually saying is it will take about 30 hrs to lower 1 K- if 6,580,000 GigaWatts is correct.
So kinda proves you right, and here I thought you were wrong :)
Though might need more reactor than 6,580,000 GigaWatts because 30 hour to lower 1 K seems kinda wildly optimist:)
So what I am getting is ground surface temperature doesn't matter. Amount of atmosphere doesn't matter. In terms reducing power requirements.

A real thermally insulated greenhouse would help :).

Edited by gbaikie, Dec 7 2011, 06:54 AM.
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The giant space ship example · Physical theory for climate