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Conditions: there is a steel plate insulated at the bottom and facing the clear night sky, the ambient air temperature is 2°C, the cosmic microwave background radiation is 3K, and the convective heat transfer coefficient is 10 formula. My job is to calculate the temperature of the steel plate at thermal equilibrium. The steel plate doesn't have given dimensions.

I've used formula for the convective heat transfer, and formula for the radiative heat transfer. Using 0.8 as the emissivity of steel, formula yields -22.158°C which is close enough to the textbook's answer of -20.9°C. This textbook has provided wrong answers multiple times in the past, so I don't fully trust it and this answer doesn't seem right to me.

Attempts to calculate in Kelvin have resulted in even more unlikely answers. Is -20°C plausible? Or where did I go wrong?

Edited to add calculations:

formula

formula

formula

My calculator puts out: formula as the solution.

mia
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    FWIW, I think that night time isn't long enough to approach equilibrium at -20C for real objects. It takes "forever" to reach "equilibrium", but check the power flow: I think it will take a long time to get close for objects with real mass, (ignoring condensation, which limits real objects). – david Feb 06 '23 at 11:13
  • The temperature is reasonable for the given information. What's missing is the atmosphere has an equivalent temperature of minus something, so the cooling is not as great. I get a value close to -17.4 C using your input. If the air temperature is -2 C instead of +2 C, then the calculation gives -20.5 C for the answer. Please show your work and someone can point out what is wrong with your calculation. – JohnHoltz Feb 06 '23 at 18:02
  • @david Yeah, I'm having an issue even visualising this scenario in its given conditions, even with all other real factors (like condensation) ignored. Thank you for your input! – mia Feb 09 '23 at 01:58
  • @JohnHoltz I've edited to add my work. The textbook definitely gives +2 C as a condition. The given solution is calculated in Kelvin, however when I put it into my calculator, it gives me completely unreasonable results (around +53 C). Could you show me how you got -17.4 C? – mia Feb 09 '23 at 02:03
  • Radiation heat transfer has to use Kelvin, not Celsius, temperature. Therefore, the equation for Qr is incorrect. – JohnHoltz Feb 09 '23 at 16:47
  • @JohnHoltz Yeah, I also tried to calculate it using Kelvin, but I got 53.88 C which doesn't seem right. – mia Feb 10 '23 at 04:54
  • I think it is impossible in real life at least for the steel plate to fall below the ambient air temp. I'd challenge you to perform or find evidence of any experiment showing otherwise. Even if it started colder, the air would warm it up. – RC_23 Feb 10 '23 at 20:45
  • @RC_23. I do not have an example for metal, but a common everyday experience (if you live in a cold climate) is frost on windshields and the ground. It can occur when the temperature is above freezing (0 C). The ground example is especially interesting because you often see that there is less (or no) frost underneath a tree. That's because the tree blocks the view of the sky, so there is less heat lost from the ground to the sky. The ground stays warmer and the frost does not form under the tree. (Come to think of it, I bet the hood of the car has frost too.) – JohnHoltz Feb 11 '23 at 03:07
  • It's really wild you mention that because I just saw a little frost on a car hood today when it is a few degrees above freezing, and was thinking about it in the context of this question. That's incredible! But do you know why the air itself does not act as a radiation source? Is it approximately transparent, $\epsilon = \alpha =0$ ? – RC_23 Feb 11 '23 at 05:33

1 Answers1

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I get a value closer to -17.40 C = 255.75 K using an iterative solution (manually in Excel).

$Q_C = 10*(2-(-17.40)) = 194 \, W/m^2$

$Q_r = 0.8*5.67\times10^{-8}*((-17.40+273.15)^4-(3)^4) = 194.06 \, W/m^2$

Note that convection can use either C or K but radiation must use K.

Since -17.4 C satisfies both equations, the answer in the textbook is wrong.

JohnHoltz
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