Calculating Earth's Heat Loss: How to Find the Rate and Total Loss in One Hour

AI Thread Summary
The discussion focuses on calculating the Earth's heat loss using the equation dQ/dt = σAT^4, where σ is the Stefan-Boltzmann constant. Initially, the user struggles with the equation and units, mistakenly calculating the area of the Earth instead of using one square meter. After clarifying the correct approach, they determine the rate of heat loss per square meter to be approximately 397.6 J/s. To find the total heat lost in one hour, they multiply this rate by 3600 seconds, resulting in about 1.4 x 10^6 J/h. The conversation emphasizes the importance of using the correct temperature scale and area for accurate calculations.
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Homework Statement



On a clear night the Earth loses heat according the equation \frac{dQ}{dt} = \sigmaAT4 If the average
temperature of the ground is 10°C, calculate the rate of heat loss, per square metre, by the Earth
and the total heat lost, per square metre, in one hour.

Homework Equations



\frac{dQ}{dt} = AT4

\sigma = 5.7 x 10-8 W m-2 K-4


The Attempt at a Solution


Well I'm not sure how you would do this? I've never been taught this equation and haven't found much information on it via google. So anyways what i did was assuming A was the surface area I found the area using 4\pir2 = 4 x pi x 6400,000^2 = 5.15x1014

Then Subbed the values into the equation: 5.15x1014 x (10)4 x 5.7 x 10-8 = 2.9 x 1011

I'm assuming this is the rate of heat loss, per square metre because its from a differential equation which is to do with rate of change...but then how do i work out the total heat loss, per square metre, per hour? Would i just times it by 60? :S

Thank You.
 
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Think about the units of sigma. What is W and what does it mean?
Also, what units of T should you be using?
 
The area to be considered is 1 square meter ("per square meter"). What scale should be used for temperature (take a look at the units on ##\sigma##).
 
Oh crap, didnt see kelvin! :/

and is the W watts? Therefore its 1 kgm^2/s^3 so then that's square meter per second cubed? :S
 
So my new answer is

5.15x10^14 x 283^4 x 5.7 x 10^-8 = 1.88x10^17

Is that the rate of heat loss per metre square? :/
 
kingstar said:
So my new answer is

5.15x10^14 x 283^4 x 5.7 x 10^-8 = 1.88x10^17

Is that the rate of heat loss per metre square? :/

That's not a square meter.

A watt is also a joule/second.
 
4 x pi x (6400,000m)^2 = 5.15x10^14 m^2?

\frac{J}{s^2m^2K^4}xK4xm2

So once that is done are you just left with J/s^2? :S
 
You're not asked for the heat loss rate from the entire surface area of the planet; just 1 square meter of it.

The result should be in units of energy/time, or watts (J/s).
 
Oh so would the area 4pi(0.5)^2 which is just pi...

Then sub that into the equation to get 1148 J/s

and then to find the total in an hour i would just do 1148 x 3600 =4.1x10^6 J/h

Is that it?
 
  • #10
Is ##4 \pi 0.5^2## the same as one square meter? What's with the ##\pi## factor? This is a flat one square meter, not a sphere.
 
  • #11
Woops, I'm making such stupid mistakes. The area of the square metre would just be 1, so put that into the equation 397.6 J/s

397 x 3600 = 1.4 x10^6 j/h

Is that it?
 
  • #12
Yes, that looks better.
 
  • #13
Thanks for the help! :D
 
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