Calculating Energy Output of the Sun on Washington State

[jpartdq]

Homework Statement

The sun outputs about 3.9x1026 W of power, a small fraction of which reaches the Earth’s surface. The sun provides about 100 W/m2 of power to the Earth’s surface averaged over time and weather. How much energy is deposited on average by the sun on Washington state in an average day? Washington is 71,302 square miles in area. Give your answer in both Joules and in kilowatt-hours (1 KWH = 3.6x106J).

Homework Equations

This I don't know. We're learning about the change in KE and PE... equations such as MGH.

The Attempt at a Solution

I don't even know where to start here. I'm not expecting an answer but can someone put me in the right direction to help me get started? I can try and figure it out from there. Maybe help me with the relevant equation... thanks!

 

[Delphi51]

You are given how much power is delivered to each square meter.
How many square meters in Washington?
The question asks for "energy"; you'll need a formula to relate energy to power.

 

[latrocinia]

W = J/s so the sun gives 100J/s /m^2 to the earth

1 mile = 1.609344 kilometers
Washinhton is 71,302 Mi^2 = 71,302 * 1,609^2 = 184592 km^2 = 184592 * 10^6 m^2there are 60*60*24 = 86400 seconds in one day

so energy output = 100 * 184592 * 10^6 * 86400 = 1594874880000 * 10^6 J = 443020800000 kWh = 4,43 * 10 ^ 11 kWh
E = J/s /m^2 * m^2 * s = J (to check for right units)

 

[Delphi51]

Looks good to me.