[{"data":1,"prerenderedAt":72},["ShallowReactive",2],{"q-fire-110-1-fire-science-001":3},{"subject":4,"subjectSlug":5,"subjectKicker":6,"subjectShort":7,"question":8,"related":26,"sameNumber":52,"hasEssay":25},"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":12,"stem":13,"options":14,"answer":19,"answerNote":20,"images":21,"imagesPending":22,"lawTimestamp":11,"explanation":23,"explanationDeep":24,"freq":12,"indexable":25},"fire-110-1-火災學概要-001","fire-110-1-fire-science-001",110,1,"下列物質各取 1 莫耳進行完全燃燒，何者所需的理論空氣量（kg）最少？",{"A":15,"B":16,"C":17,"D":18},"乙醇（CH3CH2OH）","乙烯（C2H4）","乙醛（CH3CHO）","二甲醚（CH3OCH3）","C",null,[],false,"本題考點：完全燃燒反應式之配平與理論空氣量比較——等莫耳燃料的理論空氣量,正比於完全燃燒所需之理論氧氣莫耳數。\n【正解理由】空氣中氧氣約占 21 體積%、23.2 質量%,故理論空氣量與需氧莫耳數成正比。將四者配平後,乙醛只需 2.5 莫耳氧,其餘三者均需 3 莫耳氧,乙醛所需理論空氣量最少,故選 C。\n【逐項排除】\n(A) 乙醇 C2H5OH + 3O2 → 2CO2 + 3H2O,需氧 3 莫耳;分子雖含一個氧原子,但氫數較多,抵銷後仍為 3 莫耳。\n(B) 乙烯 C2H4 + 3O2 → 2CO2 + 2H2O,需氧 3 莫耳;分子完全不含氧,所需氧全由空氣供應。\n(C) 乙醛 CH3CHO + 2.5O2 → 2CO2 + 2H2O,需氧僅 2.5 莫耳,為四者最低,理論空氣量最少,為本題應選者。\n(D) 二甲醚 CH3OCH3 + 3O2 → 2CO2 + 3H2O,與乙醇互為同分異構物,需氧同樣是 3 莫耳。\n【記憶點】同為二碳化合物,先配平再比需氧量:乙醛 2.5 莫耳最省,另三者都是 3 莫耳。","理論空氣量的求法是先配平完全燃燒反應式取得理論需氧莫耳數,再除以空氣含氧比例。以體積(莫耳)計,空氣含氧 21%,理論空氣莫耳數等於需氧莫耳數除以 0.21;要化為質量時,可取空氣平均分子量約 28.96 g\u002Fmol,或直接以空氣含氧 23.2 質量% 換算。以乙醛為例:2.5 mol × 32 g\u002Fmol = 80 g 氧,80 g ÷ 0.232 ≈ 345 g 空氣;乙醇、乙烯、二甲醚皆為 3 mol × 32 g\u002Fmol = 96 g 氧,96 g ÷ 0.232 ≈ 414 g 空氣,兩者相差約兩成。\n含氧有機物有一條速算式:化學式寫成 CxHyOz 時,理論需氧莫耳數 = x + y\u002F4 − z\u002F2。乙醇與二甲醚同為 C2H6O,得 2 + 1.5 − 0.5 = 3;乙醛 C2H4O 得 2 + 1 − 0.5 = 2.5;乙烯 C2H4 得 2 + 1 = 3。用這條式子可在考場上數十秒內解完整題,不必逐式配平,分子中自帶的氧會折抵所需外部供氧,正是乙醛勝出的原因。\n最容易被設陷阱的是「基準」:本題以每莫耳為基準,若改問每公斤燃料所需理論空氣量,須再除以分子量(乙醛 44、乙醇與二甲醚 46、乙烯 28),此時乙烯反而躍升為需空氣最多者,約 14.8 kg 空氣\u002Fkg 燃料,乙醛約 7.8。讀題時務必先確認是莫耳基準或質量基準。\n延伸考點:理論空氣量乘以空氣過剩係數即為實際供氣量;燃燒生成之二氧化碳與水蒸氣量、理論燃燒溫度推估,乃至燃燒界限與化學計量濃度,都建立在同一條配平式上。",true,[27,32,36,40,44,48],{"webId":28,"stem":29,"number":30,"year":31,"session":12},"fire-109-1-fire-science-040","有關預混合火焰特性，下列敘述何者錯誤？",40,109,{"webId":33,"stem":34,"number":35,"year":11,"session":12},"fire-110-1-fire-science-002","依據建築物火災 t2 成長理論，當釋熱率（Q）達到 4 MW 時，需要 300 秒的時間，表示火災成長之速度為下列何者？",2,{"webId":37,"stem":38,"number":39,"year":31,"session":12},"fire-109-1-fire-science-039","帶電物體或其附近之接地體，有突出部分或刃狀部分時，在該等前端近旁，所出現之微弱發光放電，此現象稱為？",39,{"webId":41,"stem":42,"number":43,"year":11,"session":12},"fire-110-1-fire-science-003","依據建築物火災的特性，當進入「穩態燃燒」階段時，釋熱率（Q）與時間（t）的關係為何？",3,{"webId":45,"stem":46,"number":47,"year":31,"session":12},"fire-109-1-fire-science-038","在橡膠中混入碳黑所製成的產品可防止人體帶靜電，其防止靜電發生的方法為？",38,{"webId":49,"stem":50,"number":51,"year":11,"session":12},"fire-110-1-fire-science-004","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？",4,[53,57,60,64,68],{"webId":54,"year":55,"stem":56,"number":12},"fire-111-1-fire-science-001",111,"近年來已有水滅火器經型式認可，下列敘述何者正確？",{"webId":58,"year":31,"stem":59,"number":12},"fire-109-1-fire-science-001","已知辛烷的燃燒下限為 0.92（vol%），根據 Burgess-Wheeler 定理其燃燒熱約為多少？",{"webId":61,"year":62,"stem":63,"number":12},"fire-108-1-fire-science-001",108,"假設火場中只含一種燃料且均勻分布，起火後火勢循 t2 成長理論持續成長，且火災成長常數呈定值。已知起火後第 1 分鐘燒耗燃料 1 公斤，則再經 2 分鐘後，從起火開始起算總共燒耗若干公斤之燃料？",{"webId":65,"year":66,"stem":67,"number":12},"fire-107-1-fire-science-001",107,"在火場中會因吸入下列何種氣體，導致阻礙紅血球輸氧功能而造成窒息死亡？",{"webId":69,"year":70,"stem":71,"number":12},"fire-106-1-fire-science-001",106,"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：",1785123162219]