Convert Watts to Kelvins on Steel Plate: 850 Sunlight Spots

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SUMMARY

The discussion focuses on calculating the temperature in Kelvins of a 2" x 2" steel plate when subjected to 28,322 watts of sunlight for 5 minutes, specifically when concentrating sunlight into 850 spots. The key equation derived is W_in = W_out(T), where W_in represents incoming power and W_out(T) denotes the power lost through conduction, radiation, and reflection as a function of temperature. The challenge lies in accurately determining W_out(T), which increases with temperature, thus complicating the calculation of the final temperature.

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douglasg14b
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How might i convert the temperature in kelvins of a 1/2inch 2" x 2" steel plate when i have 28322 watts of sunlight focused on it for 5min? also if i am stating this question wrong i will be focusing 850 2" x 2" spots of sunlight on a single 2" x 2" surface how might i calculate the temperature in K?
 
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The temperature will be the result of the balance between incoming power (your sunlight), and whatever it will lose (by conduction, radiation, reflection, ...).
So the incoming power is only one element. As the lost power is usually a strong function of the temperature (the higher the temperature, the more losses there are), we can write: W_out(T). So the end temperature will be the solution of W_in = W_out(T).

The hardest part is always to figure out W_out(T), that is: the lost power as a function of temperature.
 

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