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?
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
Explanation:
Below is an attachment containing the solution.
[tex]The intensity of electromagnetic waves from the sun is 1.4kW/m2 just above the earth's atmosphere. E[/tex]
[tex]E=8.02*10^{5}J[/tex]
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
The energy received by your back is calculated as:
[tex]E=Pt=Iat\\E=(0.80)(1400W/m^{2} )(0.30*0.51m^{2} )(1.30*3600s)\\E=8.02*10^{5}J[/tex]