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The intensity of solar radiation at the top of the Earth's atmosphere is \(1340 \mathrm{W} / \mathrm{m}^{2} .\) Assuming that \(60 \%\) of the incoming solar energy reaches the Earth's surface and assuming that you absorb \(50 \%\) of the incident energy, make an order-of-magnitude estimate of the amount of solar energy you absorb in a \(60-\) min sunbath.

Short Answer

Expert verified
The amount of solar energy an individual would absorb in a 60-minute sunbath is approximately \(1447200 \, \text{J/m}²\).

Step by step solution

01

Calculate energy reaching the Earth's surface

We know that the energy of solar radiation at the top of the atmosphere is 1340 W/m², but only 60% of this energy reaches the Earth's surface. Therefore, the energy per square meter that reaches the Earth's surface is \(0.60 * 1340 \, \text{W/m}² = 804 \, \text{W/m}²\).
02

Determine energy absorbed by the individual

The individual absorbs 50% of the incident energy. This means that \(0.50 * 804 \, \text{W/m}² = 402 \, \text{W/m}²\) is absorbed by the individual.
03

Calculate energy absorbed in a 60-minute sunbath

Energy is power times time. Given the individual absorbs 402 W/m², and knowing that there are 3600 seconds in an hour, we find the total energy absorbed in a 60-minute sunbath is \(402 \, \text{W/m}² * 3600 \, \text{s} = 1447200 \, \text{J/m}²\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Solar Radiation Intensity
Solar radiation intensity serves as a measure of how much solar energy is received per unit area from the Sun. When we discuss the intensity of solar radiation at the top of the Earth's atmosphere, a commonly cited figure is around 1340 W/m², known as the solar constant. This value isn't constant in the strictest sense—it varies slightly due to Earth's elliptical orbit and solar cycles, but it serves as a good average for calculations.

When this radiation encounters Earth, various factors like atmospheric composition, cloud cover, and the angle of incidence affect how much of that energy actually reaches the surface. In the given exercise, the assumption is that 60% of the incoming energy makes it through. Therefore, the calculation takes into account these real-world conditions to provide a more accurate estimate of available solar energy for processes such as heating, electricity generation, and photosynthesis on Earth's surface.

To capture the practical significance of solar radiation intensity, one might consider solar panel installations where estimating the potential electricity generation hinges on understanding these intensity values. In the context of our exercise, it helps to estimate how much solar energy an individual can absorb during a sunbath.
Energy Conversion Calculation
The energy conversion calculation allows us to understand how radiant energy from the Sun is converted into other forms upon interaction with objects, like heating the skin during a sunbath. It demonstrates the transformation from the energy influx measured in watts per square meter (W/m²) to the total energy absorbed over a certain time, which we measure in joules (J).

The calculation starts by identifying the percentage of solar energy an individual can absorb—50% in our scenario—which is then applied to the energy reaching the Earth's surface post atmospheric filtration. Since a watt is defined as one joule per second, to find the total energy absorbed over an hour, we multiply the energy intensity absorbed per second by the number of seconds in an hour (3600).

This conversion is essential in various scientific and engineering disciplines. For example, in the design of thermal systems, understanding how much energy is converted to heat can determine the efficiency of insulation materials or the cooling capacity needed for a building. In the sunbath scenario, it gives a tangible estimate of the energy your body would absorb from the Sun over time.
Making an Order-of-Magnitude Estimate
When calculating an order-of-magnitude estimate, the goal is to understand the scale or size of a value rather than pinpointing an exact figure. The process involves rounding numbers to a power of ten, which simplifies complex calculations and provides a quick approximation that's useful for comparison and sanity checks.

Such estimates are more about perspective than precision. They help us assess if a more detailed calculation is necessary or to confirm that our results are within a reasonable range. In our exercise, while we provide somewhat detailed steps to arrive at the amount of solar energy absorbed during a sunbath, the initial figures (60% atmospheric transmission and 50% absorption rate) are themselves rounded estimations, indicative of the order-of-magnitude approach.

Order-of-magnitude estimates are widely used in fields like astronomy, physics, and engineering to grapple with vast differences in scale and to construct initial models or hypotheses. In everyday situations, this kind of estimation equips us to make quick decisions and to understand the potential impact or significance of various phenomena or actions.

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Most popular questions from this chapter

A linearly polarized microwave of wavelength \(1.50 \mathrm{cm}\) is directed along the positive \(x\) axis. The electric field vector has a maximum value of \(175 \mathrm{V} / \mathrm{m}\) and vibrates in the \(x y\) plane. (a) Assume that the magnetic field component of the wave can be written in the form \(B=B_{\max } \sin (k x-\omega t)\) and give values for \(B_{\max }, k,\) and \(\omega\) Also, determine in which plane the magnetic field vector vibrates. (b) Calculate the average value of the Poynting vector for this wave. (c) What radiation pressure would this wave exert if it were directed at normal incidence onto a perfectly reflecting sheet? (d) What acceleration would be imparted to a \(500-\mathrm{g}\) sheet (perfectly reflecting and at normal incidence) with dimensions of \(1.00 \mathrm{m} \times 0.750 \mathrm{m} ?\)

Consider a small, spherical particle of radius \(r\) located in space a distance \(R\) from the Sun. (a) Show that the ratio \(F_{\mathrm{rad}} / F_{\mathrm{grav}}\) is proportional to \(1 / r,\) where \(F_{\mathrm{rad}}\) is the force exerted by solar radiation and \(F_{\mathrm{grav}}\) is the force of gravitational attraction. (b) The result of part (a) means that, for a sufficiently small value of \(r,\) the force exerted on the particle by solar radiation exceeds the force of gravitational attraction. Calculate the value of \(r\) for which the particle is in equilibrium under the two forces. (Assume that the particle has a perfectly absorbing surface and a mass density of \(1.50 \mathrm{g} / \mathrm{cm}^{3} .\) Let the particle be located \(3.75 \times 10^{11} \mathrm{m}\) from the Sun, and use \(214 \mathrm{W} / \mathrm{m}^{2}\) as the value of the solar intensity at that point.)

A 15.0-m \(\mathrm{W}\) helium-neon laser \((\lambda=632.8 \mathrm{nm})\) emits a beam of circular cross section with a diameter of \(2.00 \mathrm{mm}\) (a) Find the maximum electric field in the beam. (b) What total energy is contained in a \(1.00-\mathrm{m}\) length of the beam? (c) Find the momentum carried by a \(1.00-\mathrm{m}\) length of the beam.

A plane electromagnetic wave has an intensity of \(750 \mathrm{W} / \mathrm{m}^{2} .\) A flat, rectangular surface of dimensions \(50 \mathrm{cm} \times 100 \mathrm{cm}\) is placed perpendicular to the direction of the wave. The surface absorbs half of the energy and reflects half. Calculate (a) the total energy absorbed by the surface in 1.00 min and (b) the momentum absorbed in this time.

One of the weapons being considered for the "Star Wars" antimissile system is a laser that could destroy ballistic missiles. When a high-power laser is used in the Earth's atmosphere, the electric field can ionize the air, turning it into a conducting plasma that reflects the laser light. In dry air at \(0^{\circ} \mathrm{C}\) and 1 atm, electric breakdown occurs for fields with amplitudes above about \(3.00 \mathrm{MV} / \mathrm{m} .\) (a) What laser beam intensity will produce such a ficld? (b) At this maximum intensity, what power can be delivered in a cylindrical beam of diameter \(5.00 \mathrm{mm} ?\)

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