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Soil physics is the study of soil physical properties and processes. It is applied to management and prediction under natural and managed ecosystems. (Wikipedia). Keywords in this solved assignment are: Mass and Energy, Einstein, Sun's Radius, Stefan Boltzmann Law, Wien's Law, Wavelength, Soil Temerature, Wetness, Diffusion Coefficient, Moldrup relationship, Efflux
Typology: Exercises
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The sun’s radius is approximately 696,000 km, and its surface temperature is approximately 5770 K. Assuming that the sun is a blackbody, how much mass does it lose every second? Hint: you will need a fundamental relationship derived by Einstein. If the sun is a blackbody, then its emissivity ε = 1.0. The Stefan-Boltzmann law gives the energy emitted by a body as J = εσT^4 W m-2, where the constant σ = 5. x 10-8^ W m-2^ K-4. At a temperature of 5770 K, 6.3 x 10^7 J are emitted every second from 1 m^2. If the sun is a sphere with radius 696,000 km, then its surface area is 4 π (696,000 km)^2 = 6 x 10^18 m^2. Multiplying this by the energy per m^2 , the sun emits 3.8 x 10^26 J every second. Einstein gives us e = mc^2 , where c is the speed of light, approximately 3 x 10^8 m/s. The mass lost by the sun must therefore be e/c^2. So 3.8 x 10^26 J / (3 x 10^8 m/s)^2 = 4.26 x 10^9 kg lost every second – about 4.3 million tons. Incidentally, if you have trouble remembering the base units for energy, you can easily get it from Einstein’s equation. mc^2 has units of M L^2 T-2., so 1 Joule = 1 kg m^2 /s^2.
The earth’s mean temperature is approximately 288 K. Suppose that climate change raises that to 290 K: a) What is the percentage increase in the energy emitted by the earth? (Assume a constant emissivity ε = 0.95) The emissivity doesn’t affect this, since it appears in both numerator and denominator. The percentage increase is 100% * (290^4 – 288^4 )/288^4 = 2.8% b) What are the wavelengths at which maximum energy is emitted by (i) the 288 K earth, and (ii) the 290 K earth? According to Wien’s law, the wavelength at which the most energy is emitted, λλλλm, is given by λλλλm = 2900 μm K / T. So the 288 K earth has λλλλm = 10.07 μm, and the warmer earth has λλλλm = 10.00 μm, a decrease of 100% * (10.07 – 10.00) / 10.00 = 0.7%. c) What are the implications of this change in wavelength, given the absorption spectrum of the earth’s atmosphere (see, e.g., http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png)? The earth radiates heat into space, and the wavelength at which the most energy is emitted, λλλλm, happens to be a wavelength at which the atmosphere transmits readily. If the earth warmed so much that lm shifted to a wavelength at which the atmosphere did not transmit well – instead it reflected, or absorbed and re- emitted the energy – then the earth would not shed its energy readily, increasing the warming. But the shift in lm that would result from a 2 K increase in temperature should have little to no effect.
(a) The spreadsheet gives what you need to obtain the diffusion coefficient for CO 2 in free air, as a function of temperature (in °C): DCO 2 ( T) = 9.15 x 10-2^ cm^2 /s + ( T x 7.17 x 10-2^ cm^2 /s °C). Now we need to adjust this (temperature-corrected) diffusivity for the water content of the soil, at each depth and time. I recommend the Moldrup relationship: DCO 2 ( ε) = DCO 2 ε2.5^ / φ, but other relationships, like those by Penman, Marshall, or Millington, are also OK. We have θ, and we know that ε + θ = φ, but what is φ? Remember that this is field data, where we don’t have much control and have to make do with what we have. Over the days and depths give, the maximum water content is θ = 0.45. Field soils rarely exceed 90% of saturation, so let’s assume φ = 0.5. Now we can calculate DCO 2 ( θ) = DCO 2 (0.5 - θ)2.5^ / 0.5. The values I got are graphed below; yours should be pretty similar. The x-axis is day of the year, and the y-axis is diffusivity in cm^2 /s.
0 - 6 mm 6 - 12 mm 12 - 18 mm 18 - 24 mm 24 - 30 mm
Interpreting this graph isn’t the point of the homework, but: notice the daily cycle, probably due to changes in temperature because it’s strongest near the surface, and the sharp decrease around DOY 200, probably due to a significant rain event.