Question

Averaged over a year in the central United States, radiation from the Sun transfers about 200 W to each square meter of Earth's surface. If a house is 10 m long by 10 m wide, how much solar energy falls on the house each second.

Answer 1

The area of the roof of the house is (10m) x (10m) = 100 m² .

If each m² collects energy at the rate of 200 W, then the whole roof gets hit with (100 m²) x (200 W/m²) = 20,000 W .

1 watt means 1 Joule per second.

So the roof is getting broiled by 20,000 Joules of energy each second.

This is a LOT of power !  It's more than any house ever uses, even when everything inside is turned on and running. And it's FREE !  THAT's why developing efficient use of solar energy is such a big deal.  

Answer 2

The amount of solar energy that falls on the house per second is required.

The amount of solar energy that falls on the house each second is $6.31\times 10^{11}\ \text{J}$

I = Irradiance = $200\ \text{W/m}^2$

A = Area of house = $10\times 10=100\ \text{m}^2$

Power of solar radiation is

$P=IA\\\Rightarrow P=200\times 100\\\Rightarrow P=20000\ \text{W}$

Time taken to receive the given radiation is

$t=1\ \text{year}\\\Rightarrow t=1\times 365.25\times 24\times 60\times 60\\\Rightarrow t=31557600\ \text{s}$

Energy is given by

$E=Pt\\\Rightarrow E=20000\times 31557600\\\Rightarrow E=6.31\times 10^{11}\ \text{J}$

The amount of solar energy that falls on the house each second is $6.31\times 10^{11}\ \text{J}$

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