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Heat loss of the Earth

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    On a clear night the Earth loses heat according the equation [itex]\frac{dQ}{dt}[/itex] = [itex]\sigma[/itex]AT4 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.

    2. Relevant equations

    [itex]\frac{dQ}{dt}[/itex] = AT4

    [itex]\sigma[/itex] = 5.7 x 10-8 W m-2 K-4


    3. 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[itex]\pi[/itex]r2 = 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.
     
  2. jcsd
  3. May 12, 2013 #2
    Think about the units of sigma. What is W and what does it mean?
    Also, what units of T should you be using?
     
  4. May 12, 2013 #3

    gneill

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    Staff: Mentor

    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##).
     
  5. May 12, 2013 #4
    Oh crap, didnt see kelvin! :/

    and is the W watts? Therefore its 1 kgm^2/s^3 so then thats square meter per second cubed? :S
     
  6. May 12, 2013 #5
    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? :/
     
  7. May 12, 2013 #6

    gneill

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    Staff: Mentor

    That's not a square meter.

    A watt is also a joule/second.
     
  8. May 12, 2013 #7
    4 x pi x (6400,000m)^2 = 5.15x10^14 m^2?

    [itex]\frac{J}{s^2m^2K^4}[/itex]xK4xm2

    So once that is done are you just left with J/s^2? :S
     
  9. May 12, 2013 #8

    gneill

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    Staff: Mentor

    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).
     
  10. May 12, 2013 #9
    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?
     
  11. May 12, 2013 #10

    gneill

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    Staff: Mentor

    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.
     
  12. May 12, 2013 #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?
     
  13. May 12, 2013 #12

    gneill

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    Staff: Mentor

    Yes, that looks better.
     
  14. May 12, 2013 #13
    Thanks for the help! :D
     
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