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of a slab of thickness t and with thermal conductivity k is The temperature gradient, given by de = C dx where C is a constant and 0 is a function of x. By using the conditions 0(0)=01, 0()=0 2 and Fourier's law Q = -kA d0 show that (02-0,) Q = -kA- [6(0)=0, means that when x is 0 then 0301, and 0(f)=02 means that when x is t then 0(t) =02.)

of a slab of thickness t and with thermal conductivity k is The temperature gradient, given by de = C dx where C is a constant and 0 is a function of x. By using the conditions 0(0)=01, 0()=0 2 and Fourier's law Q = -kA d0 show that (02-0,) Q = -kA- [6(0)=0, means that when x is 0 then 0301, and 0(f)=02 means that when x is t then 0(t) =02.)

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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The temperature gradient, , of a slab of thickness t and with thermal conductivity k is given by de = C dx where C is a constant and 0 is a function of x. By using the conditions 0(0)=0+, 0(1)=0 2 and Fourier's law Q = -kA d0 dx show that (02-0,) Q = -kA- [(0)=0, means that when x is 0 then 0=01, and 0(t)=02 means that when x is t then 0(t) =02.]
Transcribed Image Text:The temperature gradient, , of a slab of thickness t and with thermal conductivity k is given by de = C dx where C is a constant and 0 is a function of x. By using the conditions 0(0)=0+, 0(1)=0 2 and Fourier's law Q = -kA d0 dx show that (02-0,) Q = -kA- [(0)=0, means that when x is 0 then 0=01, and 0(t)=02 means that when x is t then 0(t) =02.]
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