Yahoo Answers: Answers and Comments for Can you please help with this physics question? [Physics]
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From Connie
enUS
Mon, 27 May 2019 01:27:15 +0000
3
Yahoo Answers: Answers and Comments for Can you please help with this physics question? [Physics]
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https://answers.yahoo.com/question/index?qid=20190527012715AAc1RXH
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From az_lender: The log's surface area is
2*pi*(57 cm)*(9 ...
https://answers.yahoo.com/question/index?qid=20190527012715AAc1RXH
https://answers.yahoo.com/question/index?qid=20190527012715AAc1RXH
Mon, 27 May 2019 01:43:17 +0000
The log's surface area is
2*pi*(57 cm)*(9 cm) + 2*pi*(9 cm)^2
= 2*pi*(66 cm)*(9 cm) = 3732 cm^2 = 3.73 m^2.
33 kW/(3.73 m^2) = sigma*T^4,
where T is the absolute temperature and
sigma = 5.67 x 10^(8) W/(m^2 K^4).
T^4 = 33000 K^4/(3.73*5.67 x 10^(8))
= 1560 x 10^8 K^4 =>
T = 628K, around 355C.