The earth receives at its surface radiation from the sun at the rate of 1400 w/m2. the distance of centre of sun from the surface of earth is 1.5 × 108 m and the radius of sun is 7.0 × 108 m. what is approximately the surface temperature of the sun treating the sun as a black body?
Answers
Answered by
2
Energy Emitted = Epsilon x Sigma x T4
Epsilon = emissivity of the body emitting radiation
Sigma = Stefan Boltzmann constant = 5.67 x 10-8 W m-2 K-4
T = Temperature (K)
So we have:
Epsilon =1 - because it is a black body
T = 6000 K
Energy emitted = [1] [5.67 x 10-8 W m-2 K-4][ 6000 K]4
= 73,483,200 W m-2
If you don't know how to do this on your calculator make certain that you talked to me or one of the AIs
2. Assuming the Earth is a black body and has a temperature of 300 K, how much energy is it emitting?
Energy emitted = [1] [5.67 x 10-8 W m-2 K-4][ 300 K]4
= 459 W m-2
3. What is the peak energy emission wavelength of the Sun assuming a temperature of 6000 K? What part of the electromagnetic spectrum is this in?
To answer this question we need to use Wiens Law
I will write lambda in words rather than greek symbols
lambda max = [2.898 x 10-3 m K] / [ T]
lambda - wavelength (m or micrometers)
T - Temperature (K)
Therefore:
lambda max = [2.898 x 10-3 m K] / [ 6000 K]
= 4.83 x 10-7 m
= 0.483 micrometers
Remember that micrometers are 10-6 m
Note on P36 in your textbook there is a typo - there should be a m after the micro symbols
What part of the electromagnetic spectrum is this in?
This is in the visible part of the spectrum - see Fig 2.8 LT
4. What is the peak energy emission wavelength of the Earth assuming a temperature of 300 K? What part of the electromagnetic spectrum is this in?
lambda max = [2.898 x 10-3 m K] / [ 300 K]
= 9.66 x 10-6 m
= 9.66 micrometers
This is the infra-red portion of the spectrum
5. Assuming that a beaker of water is boiling and has a emissivity of 0.95 how much energy is it emitting? What is the peak energy emission wavelength? What part of the electromagnetic spectrum is this in?
Energy Emitted = Epsilon x Sigma x T4
= 0.95 x [5.67 x 10-8 W m-2 K-4] x [100 +273 K]4
= 1042 W m-2
Peak Energy Emission = lambda max = [2.898 x 10-3 m K] / [ 373 K]
= 7.77 micrometers
This is in the Infra red part of the spectrum
6. If the incoming solar radiation on a clear summer's day is 1000 W m-2 and the albedo over the grass surface is 0.26, the observed sky temperature is -40 deg C and the surface temperature is 25 deg C what is
a) Net shortwave radiation?
b) Net Longwave radiation?
c) Net all-wave radiation?
Assume an emissivity of 1 for the sky and 0.95 for the grass.
Note this problem is very similar to what you are going to do in Lab. You will take each of these measurements and then do the same calculations.
a) Net shortwave radiation?
K* = K dn - Kup
This time I don't have the arrows! dn - down up - up or greek letters
albedo = Kup/ Kdn
Kup = albedo x Kdn
K* = K dn - (albedo x Kdn) = 1000 - (0.26 x 1000)
= 1000 - 260
= 740 W m-2
b) Net Longwave radiation?
L* = L dn - Lup
We use Stefan Boltzmann's Law to calculate Ldn and Lup
For Ldn we use the sky temperature
Ldn=Epsilon x Sigma x T4
= [1] [5.67 x 10-8 W m-2 K-4] x [-40 +273 K]4
= 167 W m-2
Lup ==Epsilon x Sigma x T4
= [0.95] [5.67 x 10-8 W m-2 K-4] x [25 +273 K]4
= 423 W m-2
L* = Ldn - Lup
= 167 - 423 = - 255 W m-2
Note that this is negative because the sky temperature is colder than the grass temperature. When you do your observations in Lab look to see which is colder and therefore whether it will be positive or negative.
c) Net all-wave radiation?
Q* = K* + L* = (Kdn - Kup) + (Ldn - Lup)
= (740 ) + (-255)
= 484 W m-2
7. If it is a cloudy day in the winter with snow on the ground what would be the net all wave radiation under the following conditions: the albedo is 0.6 (the snow has been around for a few days and got dirty) and outgoing shortwave radiation is 300 W m-2, the surface temperature is 0 deg C and the sky temperature is -35 deg C. Assume the snow is a black body and the cloudy sky emissivity is 0.99.
Q* =[Kdn - Kup] + [ Ldn - Lup]
= [Kup/albedo - Kup] + [Epsilon x Sigma x Tsky4] - [Epsilon x Sigma x Tsnow4]
= [300/0.6 - 300] + ({[0.99] [5.67 x 10-8W m-2 K-4] x [-35 +273 K]4} -{[1] [5.67 x 10-8 W m-2 K-4] x [0 +273 K]4})
= [500 -300]+[180 - 315]
= 65 W m-2
Epsilon = emissivity of the body emitting radiation
Sigma = Stefan Boltzmann constant = 5.67 x 10-8 W m-2 K-4
T = Temperature (K)
So we have:
Epsilon =1 - because it is a black body
T = 6000 K
Energy emitted = [1] [5.67 x 10-8 W m-2 K-4][ 6000 K]4
= 73,483,200 W m-2
If you don't know how to do this on your calculator make certain that you talked to me or one of the AIs
2. Assuming the Earth is a black body and has a temperature of 300 K, how much energy is it emitting?
Energy emitted = [1] [5.67 x 10-8 W m-2 K-4][ 300 K]4
= 459 W m-2
3. What is the peak energy emission wavelength of the Sun assuming a temperature of 6000 K? What part of the electromagnetic spectrum is this in?
To answer this question we need to use Wiens Law
I will write lambda in words rather than greek symbols
lambda max = [2.898 x 10-3 m K] / [ T]
lambda - wavelength (m or micrometers)
T - Temperature (K)
Therefore:
lambda max = [2.898 x 10-3 m K] / [ 6000 K]
= 4.83 x 10-7 m
= 0.483 micrometers
Remember that micrometers are 10-6 m
Note on P36 in your textbook there is a typo - there should be a m after the micro symbols
What part of the electromagnetic spectrum is this in?
This is in the visible part of the spectrum - see Fig 2.8 LT
4. What is the peak energy emission wavelength of the Earth assuming a temperature of 300 K? What part of the electromagnetic spectrum is this in?
lambda max = [2.898 x 10-3 m K] / [ 300 K]
= 9.66 x 10-6 m
= 9.66 micrometers
This is the infra-red portion of the spectrum
5. Assuming that a beaker of water is boiling and has a emissivity of 0.95 how much energy is it emitting? What is the peak energy emission wavelength? What part of the electromagnetic spectrum is this in?
Energy Emitted = Epsilon x Sigma x T4
= 0.95 x [5.67 x 10-8 W m-2 K-4] x [100 +273 K]4
= 1042 W m-2
Peak Energy Emission = lambda max = [2.898 x 10-3 m K] / [ 373 K]
= 7.77 micrometers
This is in the Infra red part of the spectrum
6. If the incoming solar radiation on a clear summer's day is 1000 W m-2 and the albedo over the grass surface is 0.26, the observed sky temperature is -40 deg C and the surface temperature is 25 deg C what is
a) Net shortwave radiation?
b) Net Longwave radiation?
c) Net all-wave radiation?
Assume an emissivity of 1 for the sky and 0.95 for the grass.
Note this problem is very similar to what you are going to do in Lab. You will take each of these measurements and then do the same calculations.
a) Net shortwave radiation?
K* = K dn - Kup
This time I don't have the arrows! dn - down up - up or greek letters
albedo = Kup/ Kdn
Kup = albedo x Kdn
K* = K dn - (albedo x Kdn) = 1000 - (0.26 x 1000)
= 1000 - 260
= 740 W m-2
b) Net Longwave radiation?
L* = L dn - Lup
We use Stefan Boltzmann's Law to calculate Ldn and Lup
For Ldn we use the sky temperature
Ldn=Epsilon x Sigma x T4
= [1] [5.67 x 10-8 W m-2 K-4] x [-40 +273 K]4
= 167 W m-2
Lup ==Epsilon x Sigma x T4
= [0.95] [5.67 x 10-8 W m-2 K-4] x [25 +273 K]4
= 423 W m-2
L* = Ldn - Lup
= 167 - 423 = - 255 W m-2
Note that this is negative because the sky temperature is colder than the grass temperature. When you do your observations in Lab look to see which is colder and therefore whether it will be positive or negative.
c) Net all-wave radiation?
Q* = K* + L* = (Kdn - Kup) + (Ldn - Lup)
= (740 ) + (-255)
= 484 W m-2
7. If it is a cloudy day in the winter with snow on the ground what would be the net all wave radiation under the following conditions: the albedo is 0.6 (the snow has been around for a few days and got dirty) and outgoing shortwave radiation is 300 W m-2, the surface temperature is 0 deg C and the sky temperature is -35 deg C. Assume the snow is a black body and the cloudy sky emissivity is 0.99.
Q* =[Kdn - Kup] + [ Ldn - Lup]
= [Kup/albedo - Kup] + [Epsilon x Sigma x Tsky4] - [Epsilon x Sigma x Tsnow4]
= [300/0.6 - 300] + ({[0.99] [5.67 x 10-8W m-2 K-4] x [-35 +273 K]4} -{[1] [5.67 x 10-8 W m-2 K-4] x [0 +273 K]4})
= [500 -300]+[180 - 315]
= 65 W m-2
Similar questions