Ammonia, NH3, has a normal boiling point of -33 °C, and its heat of vaporization is 25.1 kJ/mol. The liquid and gas heat capacities are 2.2 J/(g °C) and 4.7 J/(g °C), respectively. How much heat must be supplied to 236 g of liquid ammonia to raise its temperature from -60.0 °C to 5.0 °C? (Hint: Draw heating curve T versus heat added for such problems).
help me please :(
Vapor Pressure problem求解
2013-11-11 12:01 am
回答 (2)
2013-11-11 4:54 am
✔ 最佳答案
no. of mole of 236 g NH₃ = 236/(14+1*3) = 13.88 moleHeat required from -60⁰C to -33⁰C :
= 2.2*236*[(-33)-(-60)] = 14018.4 J
Heat of vaporization :
= 25.1*1000*13.88 = 348388 J
Heat required from -33⁰C to -5⁰C :
= 4.7*236*[(-5)-(-33)] = 31058 J
Total amount of heat required = 14018.4 J + 348388 J + 31058 J
= 393 kJ
2013-11-11 1:57 am
氨，NH 3，有一個正常的沸點為-33℃，其汽化熱25.1千焦/摩爾。液體和氣體的熱容量為2.2 J / J /（G°C）和4.7（G°C）。多少熱量必須提供液氨236克，以提高其溫度-60.0°C至5.0°C？ （提示：繪製加熱曲線T與熱添加等問題）。
收錄日期: 2021-04-27 20:37:14
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