Question

Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take for 1.8×109J1.8×109J to arrive on an area of 1.00m21.00⁢m2?

Answer 1

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT   ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

$T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)} )\\T=1.38*10^{6} seconds$