20.厚度為 1.8 m 之煙層,測得有 60 %之光穿過,試估算其消光係數?
(A) 0.111 m-1
(B) 0.123 m-1
(C) 0.225 m-1
(D) 0.284 m-1
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統計: A(103), B(139), C(184), D(250), E(0) #3081640
統計: A(103), B(139), C(184), D(250), E(0) #3081640
詳解 (共 2 筆)
#5996924
光學密度 D=2-log(100-S)
K:消光係數(m¯1)
L:煙層厚度(m)
S:減光率
D = K * L / 2.3
依題目: 厚度為 1.8 m 之煙層 (L=1.8 m),測得有 60 %之光穿過(S減光率 = 100% - 60% = 40%),試估算其消光係數?
L=1.8 m, S減光率 = 40% 求消光係數K <牢記 log 2=0.3010, log 3=0.4771, log 7=0.8451> log相乘->相加 log相除->相減>
D=2-log(100-S)
= 2 - log (100-40)
= 2 - log 60
= 2 - log (2*3*10)
= 2 - (log 2 + log 3 + log 10)
= 2 - (0.3010 + 0.4771 + 1 )
D = 0.2219
D = K * L / 2.3 -> K = 2.3 * D / L
K = 2.3 * 0.2219 / 1.8
= 0.2835 m-1 (最接近0.284)
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