Remote sensing, Question from an old exam

[Firben]

Homework Statement

A satellite infrared radiometer (λ = 10 micrometer, footprint size 5x5 km) is observing a region of mixed open water and ice covered sea surface, where the temperature of the open water is -1 degree celsius and that of the sea ice -7 degree celsius. In the infrared, sea ice can be assumed as blackbody, while open water has an emissivity of about 0.8. Neglect any interference of the atmosphere

What is the brightness temperature if the pixel contains 30 % pure open water and 70 % sea ice ?

Homework Equations

Tb^-1 = k/(h*v)*ln[1+e^((h)(v)/((k)(T))-1)/ε]

ε = 1 for a black body

The Attempt at a Solution

i got the observed brighness temperature for pure water to be 261.049 K and 266.0000846 K for pure sea ice
(Expected value)
E(ξ) = 0.3*261.049 + 0.7*266.0000846 = 264.514 K

This is not valied since it is Rayleigh-Jeans , Why ?

It should by Planck but i don't know the intensity ?

 

[haruspex]

I know nothing about this subject, but it seems wrong to me to take the weighted average of the temperatures. The detector does not directly measure temperature, it measures total power received at a certain wavelength. Shouldn't you take the weighted average of those powers for the two sources, then infer the apparent temperature?

 

[Firben]

Why must i use the Planck function ?

 

[Firben]

If i start by calculating the brightness temperature of open water (grey body) by using this equation:

https://wikimedia.org/api/rest_v1/media/math/render/svg/c993a06e73c39413e249b260d33368395a75b5d4
(Tb^-1 = k/(h*v)*ln[1+e^((h)(v)/((k)(T))-1)/ε])

and plug in my values, i get the following temperature for open water T ~ 262.89 K

Then when i calculate the brightness temperature for sea water (which is a black body). I used Planck's law:

https://wikimedia.org/api/rest_v1/media/math/render/svg/5a1df6cc98f89d1f11d7eb5425cc67b4dd9bd8a2
(Planck's law)

for the intensity. And for the brightness temperature i used the equation:

https://wikimedia.org/api/rest_v1/media/math/render/svg/0ffad7df33ab06c82841a0030ed0c67abaefcbff

Tb = (hc/kλ)ln^-1(1+ 2hc^2/(I*λ^5))

(For some reason my equation's can't be seen here, they are from wikipedia: url:https://en.wikipedia.org/wiki/Brightness_temperature#Calculating_by_wavelength)

Now when the pixel contain 70 % sea ice, I multiplied the Planck's law with 0.7 and plugged in my values in the equation below and got the brightness temperature to be equal to T ~ 57 K. Which looks way to cold. If i average the sea water and the open water that the pixel contain: (57 K + 262.89 K)/2 i get the following value T(avg) ~ 160 K. I think i have missed something here ? its look's to cold.

 

[Firben]

Is this the right method ?

 

[haruspex]

Is this the right method ?

As I wrote, it does not seem right to me to be averaging temperatures. The sensor does not directly sense those.
For each of the water bodies, you can calculate the intensity at the given wavelength. The sensor will sense the weighted average of these. You can then turn that intensity back into a temperature.

The equation at the Wikipedia site doesn't make sense to me. The -1 exponent on the ln() looks like an error. If I derive it from the preceding equation (I_λ=) I get just plain ln().

 

[Firben]

If i start by calculate the intensity of the sea ice and the open water, and then take the weighted average of the two intensities. I then got a intensity value of 2.155*10^10 w/m^2, in which i put into the brightness temperature equation. And from there i got a brightness temperature value of 267.403 K.

 

[haruspex]

If i start by calculate the intensity of the sea ice and the open water, and then take the weighted average of the two intensities. I then got a intensity value of 2.155*10^10 w/m^2, in which i put into the brightness temperature equation. And from there i got a brightness temperature value of 267.403 K.

Those intensities are at the sensors wavelength, right?
Without checking your arithmetic, that seems a reasonable answer.