# Temperature of the earth

1. The Earth intercepts 1.27X10^17W of radiant energy from the Sun. Suppose the Earth, of volume 1.08X10^27 m^3, was composed of water. How long would it take for the Earth at 20 celciues to reach 90 celcius , if none of the energy was radiated or reflected back out into space?
The specific heat of water is 4.186J/g c /b]

a)154.8 y
b)7.46X10^4 y
c)7.9X0^5 y
d)7.24X10^7 y
e)78 y

3.can't find an equation that has time radiation and temperature difference.:s/b]

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Matterwave
Gold Member
If the water doesn't radiate or reflect the heat away, it is simply just absorbing all that energy right? So how much energy does it absorb in 1 second? (Hint: Watts are Joules/second) How much energy is required to heat up an Earth's volume worth of water from 20 to 90 degrees C? (Hint: What is the density of water, and how are density and volume related to mass? How much energy does it take to heat 1 kg of water from 20-90 degrees C?)

ok.i was in a wrong direction i get this now but still,
so power given=1.27*10^17W=energy/time
then
mc delta(t)=energy
densityXvolumeXdelta(temp)=energy
1000x1.08X10^27X4186X70=energy=7.56X10^31
then
time=E/W=2.49X10^18sec
years=time/31536000=7.9x10^10years:s
not an answer in the choices!

Matterwave
Gold Member
Are you sure choice c isn't 7.9*10^10? I get 7.9*10^10 too, I don't know why it should be different.

http://img535.imageshack.us/img535/2184/70212411.png [Broken]

This is the question as is :s

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any idea?

sylas
any idea?
Yes. The question gives the volume of the Earth incorrectly. It should be 1.08*1021 m3. This reduces things by 106.

Hence the answer is 7.9*104 years.

Still not in their list. Someone start from scratch without uising any of their numbers... I have to go back to finish watching a TV program. :tongue:

edit. commercial break. They also have solar energy intercepted wrong. They seem to be counting solar energy absorbed, by not counting the 30% reflected by Earth's albedo. So they should either use about 1.76*10^17 "intercepted", or call it energy absorbed. Back in a tick...

edit again. OK. The question explicitly says "if none of the energy was radiated or reflected back into space". There are all kinds of problems there also; it's physically impossible. As soon as water has any kind of temperature it will radiate. But that's okay, they set the parameters. Solar constant at Earth's orbit is about 1365 W/m2. Cross section area, which indicates how much is actually intercepted, is pi.R2. R is about 6.4*106 m. So the energy ought to be about 1.76*1017 W.

edit again. Given all these revised numbers, I get 57,000 years.

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this is cengage i dont think they will have wrong choices :s

D H
Staff Emeritus
For shame. Somebody didn't pay attention to units -- and lost a factor of 10 to boot. Using the density of water is 1, and ignoring units,

1.08e27 * 1 * 4.186 * 70 / 1.27e17 / 365.25 / 86400 = 7.9e4

The correct answer is 7.9×1010 years -- given the parameters of the question, of course.

so i should answer 7.9x10^5 in order to get the answer they think is correct
?although x10^10 is the correct one?

D H
Staff Emeritus
That the Earth's volume is stated as 1.08×1027 m3 as opposed to 1.08×1027 cm3 is yet another units mistake -- as is using the specific heat of water as 4.186 J/g/C° (which mixes SI and cgs units).

even if they mistook and calculated density as 1 instead of 1000 it makes it X10^4 as you say which is still not an answer:s im confused

so it's definetaly 7.9X sth so i should just go with the only 7.9 answer X10^5.I have another try at the assignment .and the numbers will change so i can get correct numbers then.Thanks for everybody for their help.I appreciate it :)

sylas
even if they mistook and calculated density as 1 instead of 1000 it makes it X10^4 as you say which is still not an answer:s im confused
Ignore the numbers they have given. Let's start from scratch, using the following, all in SI units.

• The solar constant is 1365 W/m2
• The Earth radius is 6.38 * 106 m. (I'll ignore the fact that the equatorial and polar radius are slightly different).
• The cross-section area of the Earth to the Sun is pi.R2, or 1.28*1014 m2
• The solar energy intersecting the Earth is thus 1.75*1017 W.
• The volume of the Earth is 4/3.pi.R^3, or 1.09*1021 m3. (Actually, 1.08 is closer, given the slightly flattened shape of the geoid, but hey.)
• The density of water is 1000 kg/m3.
• Hence, the mass of the Earth's volume as water is 1.09*1024 kg. (The actual mass of the Earth is nearly 6*1024 kg, if anyone is interested.)
• The specific heat of water is 4186 J/kg/K
• Therefore it takes 4.55*1027 J to raise this mass of water by 1 kelvin.
• 20 to 90 C is 70 K.
• Therefore the energy to raise that volume of water from 20 to 90 C is 3.19*1029 J.
• If all the energy from the Sun goes into heating the water, with no losses, then the time would be 1.83*1012 s.
• A year is 3.156*107 s.
• Final answer, therefore, is 5.79*104 years.

well this answer is still not in the choices
wht to choose then?

sylas
well this answer is still not in the choices
wht to choose then?
Any choice you make from what is given will be wrong.

I'm simply trying to figure what the answer is, under the stated assumption that all the solar energy impacting the Earth goes into heating the water, with nothing radiated or reflected.

Cheers -- sylas

I have to make one of the given choices this is the online assignment from cengage .ill try to tell my teacher abt the mistake but i must make a choice.what is best choice.ur given energy that u calculated from correct givens divided by the power the question assumed gived a close one 7.9X10^4 years .i donno wat to choose.Thanks anyway for all the help.:):)

sylas