[{"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,[53,57,61,65,69],{"webId":54,"year":55,"stem":56,"number":12},"fire-114-1-fire-science-001",114,"火場中輻射為熱傳遞主要的方式之一，輻射熱之性質及其計算的方式下列何者錯誤？",{"webId":58,"year":59,"stem":60,"number":12},"fire-113-1-fire-science-001",113,"有一倉庫長寬高分別為 8 公尺，5 公尺及 4 公尺，內置可燃物之總重量為 600 公斤，試計算該倉庫之火載量若干（kg\u002Fm2）？",{"webId":62,"year":63,"stem":64,"number":12},"fire-112-1-fire-science-001",112,"建築物火災造成人命傷亡的原因，主要是煙與有毒氣體，下列何者為防煙最好的方法？",{"webId":66,"year":67,"stem":68,"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":70,"year":31,"stem":71,"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？",1785129966932]