Thermal Radiation Homework: Snow-Covered Surface Absorption

AI Thread Summary
To determine how much solar radiation is absorbed by a snow-covered surface, start with the total solar radiation of 1367 W/m², noting that 50% is absorbed or reflected before reaching the surface, leaving 683.5 W/m². The emissivity of snow is crucial for calculations, typically ranging from 0.969 to 0.997. This value affects the thermal radiation emitted by the snow, which can be calculated using the equation Hr = (Area)(e)(5.67x10^-8 W/m²K⁴)(temperature)⁴. Understanding these concepts is essential for solving the homework problem effectively.
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Homework Statement



Consider a snow-covered surface. Assume that the amount of solar radiation that reaches the top of the atmosphere is 1367 Wm-2. Given that 50% of this solar radiation is either absorbed in the atmosphere or reflected by clouds before reaching the the snow covered surface, how much of the original shortwave radiation is absorbed by the snow?

Homework Equations


Hr= (Area)(e)(5.67x10^-8 W/m^2K^4)(temperature)^4

The Attempt at a Solution


I seriously don't know how to.. please help me..
 
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What is "emissivity" of snow?
 
Bystander said:
What is "emissivity" of snow?

I searched it. it says that it ranges from 0.969 - 0.997
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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