# The intensity of electromagnetic waves from the sun is 1.4kW/m2 just above the earth’s atmosphere. Eighty

The intensity of electromagnetic waves from the sun is 1.4kW/m2 just above the earth's atmosphere. Eighty percent of this reaches the surface at noon on a clear summer day. Suppose you think of your back as a 26.0 cm × 52.0 cm rectangle. How many joules of solar energy fall on your back as you work on your tan for 0.90 h?

## This Post Has 3 Comments

1. lovebug7685 says:

The answer is 4.9 × 10^5 J

Explanation:

Recall that,

Power (P) = Energy (E) / time (t)

Hence,

Energy (E) = Power (P) × time (t)

First, we will determine the power, P

From, Intensity (I) = Power (P) / Area (A)

Power is then given by

Power (P) = Intensity (I) × Area (A)

From the question, the intensity of electromagnetic waves from the sun is 1.4kW/m2, but only 80% of this reaches the surface at noon.

Hence,

Intensity (I) = 80% of 1.4kW/m2

I = 0.8 × 1.4 × 10^3 W/m2

I = 1120 W/m2

For the area,

Area of rectangle (A) = length × breadth

A = 26.0 × 10^-2 m × 52.0 × 10^-2 m

A = 0.1352 m2

Hence,

Power (P) = 1120 W/m2 × 0.1352 m2

P = 151.424 W

Now, to the energy,

From,

Energy (E) = Power (P) × time (t)

P = 151.424 W

From the question,

t = 0.90 h = (0.9 × 60 × 60) s = 3240s

Hence,

E = 151.424 × 3240

E = 490613.76 J

E = 4.9 × 10^5 J

2. jairopanda8 says:

Explanation:

Below is an attachment containing the solution.

$The intensity of electromagnetic waves from the sun is 1.4kW/m2 just above the earth's atmosphere. E$

3. ravenalionna115 says:

$E=8.02*10^{5}J$

Explanation:

Given data

Electromagnetic waves from the sun is I=1.4kW/m² at 80%

Area a=(0.30×0.51)m²

Time t=1.30 hr

To find

Energy E

Solution

$E=Pt=Iat\\E=(0.80)(1400W/m^{2} )(0.30*0.51m^{2} )(1.30*3600s)\\E=8.02*10^{5}J$