Re: Um. Continuity?


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Re: Um. Continuity?

Posted By: thermoplyae Date: 9/2/06 10:07 a.m.
In Response To: Um. Continuity? *NM* (Document)

: Also, could bacteria really live in a body floating space? I mean, maybe they
: might be insulated if they're deep enough inside a tissue, but it seems
: like the whole thing would cool to background temperature pretty quickly.
Radiative heat transfer of a body is given by H = A * e * sigma * T⁴, where H is dQ/dt (heat transfer per second), A is the body's surface area, sigma is a Stefan-Boltzmann constant (5.67x10^-8 W m^-2 K^-4), and T is the temperature of the body in degrees Kelvin. Heat itself is related to temperature by Q = m C T, where Q is heat, m is the mass of the body, C is its specific heat, and T is its temperature. Substitution yields:
^dQ/_dt = -A e sigma (^Q/_mC)⁴
^dQ/_dt = -k Q⁴ for k = ^{(A e sigma)} / _{(m⁴ C⁴)}
The solution from here is trivial.
This might not have been the answer you were looking for, but it was the best I could do.

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: Radiative heat transfer of a body is given by H = A * e * sigma * T 4 , where : H is dQ/dt (heat transfer per second), A is the body's surface area, sigma : is a Stefan-Boltzmann constant (5.67x10 -8 W m -2 K -4 ), and T is the : temperature of the body in degrees Kelvin. Heat itself is related to : temperature by Q = m C T, where Q is heat, m is the mass of the body, C is : its specific heat, and T is its temperature. Substitution yields: dQ / dt : = -A e sigma ( Q / mC ) 4 : dQ / dt = -k Q 4 for k = (A e sigma) / (m 4 C 4 ) : The solution from here is trivial. : This might not have been the answer you were looking for, but it was the best : I could do.

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I had (have) utterly too much time on my hands...	LegacyTyphoon	8/27/06 3:05 a.m.
Re: I had (have) utterly too much time on my hands	Adam Ashwell	8/27/06 6:48 a.m.
Re: I had (have) utterly too much time on my hands	Tyler	8/27/06 11:58 a.m.
Re: I had (have) utterly too much time on my hands	LegacyTyphoon	8/27/06 6:46 p.m.
Re: I had (have) utterly too much time on my hands	Alexander Smith	8/31/06 5:36 p.m.
Um. Continuity? NM	Document	8/31/06 5:58 p.m.
Re: Um. Continuity?	thermoplyae	9/2/06 10:07 a.m.
What? NM	Alexander Smith	9/1/06 4:45 a.m.
Re: I had (have) utterly too much time on my hands	LegacyTyphoon	9/1/06 8:21 p.m.
Re: I had (have) utterly too much time on my hands	Document	9/1/06 8:35 p.m.
Re: I had (have) utterly too much time on my hands	LegacyTyphoon	9/1/06 9:19 p.m.
Re: I had (have) utterly too much time on my hands	Document	9/2/06 5:59 a.m.
Re: I had (have) utterly too much time on my hands	LegacyTyphoon	9/2/06 11:38 p.m.
Re: I had (have) utterly too much time on my hands	Nicholai	9/3/06 4:39 p.m.



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	Message: : Radiative heat transfer of a body is given by H = A * e * sigma * T 4 , where : H is dQ/dt (heat transfer per second), A is the body's surface area, sigma : is a Stefan-Boltzmann constant (5.67x10 -8 W m -2 K -4 ), and T is the : temperature of the body in degrees Kelvin. Heat itself is related to : temperature by Q = m C T, where Q is heat, m is the mass of the body, C is : its specific heat, and T is its temperature. Substitution yields: dQ / dt : = -A e sigma ( Q / mC ) 4 : dQ / dt = -k Q 4 for k = (A e sigma) / (m 4 C 4 ) : The solution from here is trivial. : This might not have been the answer you were looking for, but it was the best : I could do.
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