a car parked in sun absorbs energy at rate of 590 watts per square meter of surface area. the car reaches a temperature at which it radiates energy at the same rate. treating the car as a non perfect radiatior find the temp of the car.
e=0.8
SB constant= 5.67x10^-8
t^4=Q/t*a*e*SBC
and what is Q?
e=0.8
SB constant= 5.67x10^-8
t^4=Q/t*a*e*SBC
and what is Q?
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According to SB, Q = tAεσT⁴. We can rearrange this equation to get power emitted in J/s/m² or W/m² by dividing both sides by time t and area A:
F = εσT⁴
If the car is emitting at a rate of F=590 W/m², solving for temperature:
εσT⁴ = 590
T⁴ = 590/(εσ)
T = (590/(εσ))^(1/4)
T = 337.7 K
F = εσT⁴
If the car is emitting at a rate of F=590 W/m², solving for temperature:
εσT⁴ = 590
T⁴ = 590/(εσ)
T = (590/(εσ))^(1/4)
T = 337.7 K
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I found it at http://milanstreet.info/419849/car-park