How much heat must be added to 155 g of Ethanol at 30.0°C toconvert all the liqu

Question

How much heat must be added to 155 g of Ethanol at 30.0°C toconvert all the liquid into vapor at 100.0°C? Give your answerin J to 3 significant figures. Boiling point (K) 351.5 CP,m (liquid ethanol) (J· (°C) 1 · mol1) 111.46 CP,m (gaseous ethanol) (J· (°C) 1 · mol1) 65.44 Hvap (kJ · mol1) 43.5
What I did was converted my grams of ethanol to moles and myHvap to J/mol so that my units werethe same and then I set it up as: (3.36*111.46*48.5)+(43500*3.36)+(3.36*65.44*21.5) the answer I got to 3 significant figures was 169000 and thiswas wrong. I am wondering what I did wrong and how to go about thisproblem in the right way. Boiling point (K) 351.5 CP,m (liquid ethanol) (J· (°C) 1 · mol1) 111.46 CP,m (gaseous ethanol) (J· (°C) 1 · mol1) 65.44 Hvap (kJ · mol1) 43.5
What I did was converted my grams of ethanol to moles and myHvap to J/mol so that my units werethe same and then I set it up as: (3.36*111.46*48.5)+(43500*3.36)+(3.36*65.44*21.5) the answer I got to 3 significant figures was 169000 and thiswas wrong. I am wondering what I did wrong and how to go about thisproblem in the right way. Boiling point (K) 351.5 CP,m (liquid ethanol) (J· (°C) 1 · mol1) 111.46 CP,m (gaseous ethanol) (J· (°C) 1 · mol1) 65.44 Hvap (kJ · mol1) 43.5

Explanation / Answer

       Number of moles ofethanol = 155g / 46.07 g/mol                                                   = 3.364 mol         Boiling point ofethanol = 351.5K = 78.5oC       step 1 :  heating 3.364 moles of liquid ethanol from 30oC to78.5oC                         q1 =3.364 mol * 111.46 J/oC .mol * (78.5-30oC)                = 18185.1 J      step 2 : liquid ethanol at78.5oC to gaseous ethanol at 78.5oC                     q2 = 3.364 mol * Hvap                           = 3.364 mol * 43.5 x 103 J/mol                           = 146334 J      step 3:  heating Ethanol(g) from 78.5oC to 100oC                    q3 = 3.364 mol * 65.44 J/oC .mol *(100oC-78.5oC)                          = 4733.01 J            The total energy Q = q1 + q2 +q3                                           = 169252.11 J

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