# Heat loss of the Earth

1. May 12, 2013

### kingstar

1. The problem statement, all variables and given/known data

On a clear night the Earth loses heat according the equation $\frac{dQ}{dt}$ = $\sigma$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

$\frac{dQ}{dt}$ = AT4

$\sigma$ = 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$\pi$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. May 12, 2013

### barryj

Think about the units of sigma. What is W and what does it mean?
Also, what units of T should you be using?

3. May 12, 2013

### 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 $\sigma$).

4. May 12, 2013

### kingstar

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

5. May 12, 2013

### kingstar

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? :/

6. May 12, 2013

### Staff: Mentor

That's not a square meter.

A watt is also a joule/second.

7. May 12, 2013

### kingstar

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

8. May 12, 2013

### 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).

9. May 12, 2013

### kingstar

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. May 12, 2013

### 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.

11. May 12, 2013

### kingstar

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. May 12, 2013

### Staff: Mentor

Yes, that looks better.

13. May 12, 2013

### kingstar

Thanks for the help! :D