1. What solar temp is needed for the peak intensity of radiation to occue at .2 micrometers? Remembering that humans can see light only between .4 and .7 microns, would the sun look brighter or dimmer at this new temp?
2. How much variation in earth orbital distance from the sun is needed to alter the solar constant by 10%?
THANKS!!
Wien's displacement law
Lamdamax*temp=b
where λmax is the peak wavelength, T is the absolute temperature of the black body, and b is a constant of proportionality called Wien's displacement constant, equal to 2.8977721(26)×10−3 m K.[1]
Watch your units.
2. It changes according to the inverse square law.
1.10=(orbital radius/changed distance)^2
or orbital distance changes =sqrt1.1
= about 1.05 or five percent
dentify the fruit or vegetable you selected and relate the second law of thermodynamics to the changes in energy observed in Appendix CHow did the amount of energy gained compare to the amount of energy lost as heat at the second through fourth trophic levels? Considering the amount of energy required to produce animal-based foods and goods, should humans change their habits so they consume products closer to the bottom of the food chain because it is more efficient?
1. To determine the solar temperature needed for the peak intensity of radiation to occur at 0.2 micrometers (μm), you can use Wien's displacement law. According to this law, the wavelength of maximum intensity (λmax) is inversely proportional to the temperature (T) of a black body emitter:
λmax * T = constant
Using this formula, you can solve for the temperature required for a peak intensity at 0.2 μm. We know that the peak intensity for visible light (which the human eye can see) is around 0.55 μm. Since the Sun emits radiation in the visible spectrum, its peak intensity occurs around this wavelength.
Let's assume the temperature required for a peak intensity at 0.2 μm is Tnew. So the equation becomes:
0.2 μm * Tnew = 0.55 μm * Tsun
Where Tsun is the temperature of the Sun required for the peak intensity at 0.55 μm. Solving for Tnew gives:
Tnew = (0.55 μm * Tsun) / 0.2 μm
Now, if Tnew is greater than Tsun, it means that the Sun needs to be hotter to have a peak intensity at 0.2 μm. Therefore, at this new temperature, the Sun would look brighter to the human eye.
2. To determine the variation in Earth's orbital distance from the Sun needed to alter the solar constant by 10%, we can use the inverse square law. According to this law, the solar constant (the amount of solar radiation received at Earth's distance) is inversely proportional to the square of the distance (d) from the Sun:
Solar Constant ∝ (1 / d^2)
Let's assume the normal Earth-Sun distance is denoted as d0 and the solar constant at this distance as S0. Now, we want to find the change in distance (Δd) that would cause a 10% change in the solar constant, denoted as ΔS:
ΔS = 0.1 * S0
To find Δd, we can rearrange the equation as:
ΔS = S0 - S1 = 0.1 * S0
where S1 is the solar constant at a new distance d1, which is d0 + Δd. Using the inverse square law:
(1 / (d0 + Δd)^2) - (1 / d0^2) = 0.1
The value of Δd can then be solved using algebraic methods or numerical approximation techniques like Newton's method or bisection method.
Note: The actual numerical values used in these calculations would depend on the specific constants and units chosen, but the general methodology outlined here should provide the basis for finding the answers to both questions.