Calculating Equilibrium Temperature of Blacktop Road Under High Noon Radiation

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At high noon, a blacktop road absorbs 1,000 W/m² from the Sun. To find the equilibrium temperature, the energy absorbed must equal the energy radiated, described by the formula P = σAeT^4. The emissivity (e) for an ideal radiator is 1, which simplifies calculations. Users discuss how to rearrange the formula to solve for temperature (T) by plugging in the values. The conversation emphasizes understanding the principles of thermal equilibrium and the application of the radiation formula.
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At high noon, the Sun delivers 1,000 W to each square meter of a blacktop road. If the hot asphalt loses energy only by radiation, what is its equilibrium temperature?

I know the formula for energy rate of radiation:
P = \sigma A e T^4
I don't know how to apply this to the problem...
Any help would be great! Thx in advance ! :)
 
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If the blacktop is in thermal or radiative equilibrium it radiates the same amount of energy it absorbs.
 
So do I solve for T for the above formula and plug in the values? If so, then what is the emissivity(e)?
 
e = 1 for an ideal radiator.
 
thanks a lot! :)
 
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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