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?

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  1. 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. 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]

  3. [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]

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