[{"data":1,"prerenderedAt":70},["ShallowReactive",2],{"q-fire-112-1-fire-science-004":3},{"subject":4,"subjectSlug":5,"subjectKicker":6,"subjectShort":7,"question":8,"related":27,"sameNumber":53,"hasEssay":26},"火災學概要","fire-science","火災學 · Fire Science","火災學",{"id":9,"webId":10,"year":11,"session":12,"subject":4,"number":13,"stem":14,"options":15,"answer":20,"answerNote":21,"images":22,"imagesPending":23,"lawTimestamp":11,"explanation":24,"explanationDeep":25,"freq":12,"indexable":26},"fire-112-1-火災學概要-004","fire-112-1-fire-science-004",112,1,4,"若空氣中的氧氣體積百分比為 21%，則甲烷完全燃燒的化學理論濃度為多少%？",{"A":16,"B":17,"C":18,"D":19},"5.2","9.4","11.0","15.4","B",null,[],false,"本題考點：可燃氣體「化學理論濃度」（Cst）之計算，即燃料與空氣恰好完全燃燒、無剩餘氧氣時的燃料體積百分比。\n【正解理由】依燃燒化學計量式，甲烷完全燃燒為 CH4 + 2O2 → CO2 + 2H2O，每1莫耳甲烷需2莫耳氧；空氣中氧佔21體積百分比，故所需空氣為 2÷0.21 莫耳。化學理論濃度定義為燃料體積除以（燃料＋空氣）體積，計算得約9.48%，四捨五入後與選項9.4最接近，故選B。\n【演算步驟】\n需氧莫耳數 n = 2 mol。\n理論空氣量 = 2 mol ÷ 0.21 = 9.524 mol。\nCst = 1 mol ÷ (1 mol + 9.524 mol) × 100% = 9.5%，以通式 100 ÷ (1 + 4.773 × 2) = 9.48%，取9.4。\n【逐項排除】\n(A) 5.2接近甲烷燃燒下限約5%，是可燃範圍低端值，非恰好完全燃燒之理論濃度。\n(B) 9.4即上述計算值，位於可燃範圍中央偏下，符合理論濃度特性，正確。\n(C) 11.0來自分母漏加燃料本身1莫耳，只用1÷9.524得10.5進位而來。\n(D) 15.4接近甲烷燃燒上限約15%，屬可燃範圍高端值，與理論濃度無關。\n【記憶點】Cst = 100 ÷ (1 + 4.773n)，甲烷 n=2 得9.5%，「下限5、理論9.5、上限15」一線記牢。","化學理論濃度是火災學計算題的骨幹公式，來源是道爾頓分壓觀念下的體積比：理想氣體同溫同壓時，莫耳比即體積比，所以把配平式的莫耳數直接當體積用即可。式中的常數4.773正是1÷0.21，代表每需要1莫耳氧就得帶進4.773莫耳空氣（含約3.773莫耳氮氣當作陪伴的惰性氣體），因此通式寫成 Cst = 100 ÷ (1 + 4.773n)。只要能寫出正確的配平式取得n，任何碳氫化合物都能套：甲烷n=2得9.5%，乙烷n=3.5得5.6%，丙烷n=5得4.0%，丁烷n=6.5得3.1%，可以看出分子愈大、需氧愈多，理論濃度愈低，這個遞減趨勢本身就是選擇題的判斷捷徑。含氧或含氯的燃料則要把分子自帶的氧扣掉，如甲醇 CH3OH 的n為1.5。\n這個數值真正的用處在於與燃燒界限互相驗證。經驗上燃燒下限約為化學理論濃度的0.55倍、上限約為3.5倍（瓊斯法則），甲烷以9.5%推算得下限約5.2%、上限約33%，下限與實測5%非常吻合，上限則因實際反應動力學限制而落在15%左右，這也解釋了為何選項會同時放進5.2與15.4當誘答——它們是同一組數列上的鄰居，考生若只背「甲烷5至15」而沒理解各自意義就會誤選。\n延伸考點還有勒沙特列公式：混合氣體的燃燒下限為 100 ÷ Σ(各成分體積百分比 ÷ 各自下限)，常與本題同章出現。另外要區分理論濃度與「最易引燃濃度」，後者通常略高於理論濃度，因為稍富燃料側的火焰傳播速度最快、最小引火能量最低，這也是瓦斯外洩事故中爆炸威力最大的濃度區。把配平、常數4.773、瓊斯倍率三件事串起來，火災學的氣體計算題幾乎都能一式通解。",true,[28,33,37,41,45,49],{"webId":29,"stem":30,"number":31,"year":32,"session":12},"fire-111-1-fire-science-040","沾有油類之毛巾，集中堆置於塑膠籃而產生自然發火，其原因為下列何者？",40,111,{"webId":34,"stem":35,"number":13,"year":36,"session":12},"fire-114-1-fire-science-004","下列有關隧道火災之敘述何者錯誤？",114,{"webId":38,"stem":39,"number":40,"year":32,"session":12},"fire-111-1-fire-science-039","撕下新購的電視機螢幕保護封膜時遭受靜電，其原因為下列何者？",39,{"webId":42,"stem":43,"number":12,"year":44,"session":12},"fire-106-1-fire-science-001","某物質每克的燃燒熱為 1 仟焦耳，若 100 克該物質於 10 秒內平均燃燒完，則該物質之熱釋放率為：",106,{"webId":46,"stem":47,"number":48,"year":32,"session":12},"fire-111-1-fire-science-038","在長 10 m、寬 8 m、高 3 m 的居室燃燒 1 kg 的乳膠枕頭，其 Dm=0.65 m2\u002Fg，估算此火場出口標示燈能見度？",38,{"webId":50,"stem":51,"number":52,"year":44,"session":12},"fire-106-1-fire-science-002","造成工業火災的最高比率因素為何？",2,[54,55,58,62,66],{"webId":34,"year":36,"stem":35,"number":13},{"webId":56,"year":32,"stem":57,"number":13},"fire-111-1-fire-science-004","日本東京放映所及精神病診所火災起因為縱火，請問汽油在空氣中的燃燒下限為何？",{"webId":59,"year":60,"stem":61,"number":13},"fire-110-1-fire-science-004",110,"在氣溫 20 ℃無風的情況下，一般木造建築物形成危險界限溫度之輻射熱通量超過多少 kcal \u002F m2 h 即有延燒之危險？",{"webId":63,"year":64,"stem":65,"number":13},"fire-109-1-fire-science-004",109,"某一火災成長至 2,000 kW 所需時間為 400 秒，請問該火災屬何種成長性火災？[Q：釋熱率（MW）； 1\u002F2 t：經過時間（sec）；k：火災成長常數（sec\u002FMW ）]",{"webId":67,"year":68,"stem":69,"number":13},"fire-108-1-fire-science-004",108,"今有 A、B 兩種可燃性氣體以 1：1 之體積混合，已知 A、B 之燃燒下限分別為 3.0%、7.0%，若兩者間互不催化亦不反應，則理論上此混合氣體之燃燒下限為若干%？",1785123161448]