This is the fourth time I have attempted to have this uestioned answered. Please

Question

This is the fourth time I have attempted to have this uestioned answered. Please do not answer if you plan to heavily abbreviate or skip steps. I prefer the answer to be in Latex or in a photograph. Thank you so much. The last answer I got

total heat q required to convert 100 g of liq. MeOH at 32.60 oC to vapor at 95.60 oC = 122.1 kJ = 122.1 x 1000 J

Convert liquid MeOH at 32.60 oC to liquid MeOH at 64.60 oC

q1 = mCpdT = 100 x Cp x (64.6-32.6)

Convert liquid MeOH to vapor at 64.60 oC

q2 = mdHvap = 100 x 35.21 x 1000/32.04 = 1.10 x 10^5 J

Convert vapor MeOH at 64.60 oC to vapor at 95.60 oC

q3 = mCpdT = 100 x 0.328 x 4.1840 x (95.60-64.60) = 4254.2912 J

Add q1, q2 and q3, the total showuld be q

122.1 x 1000 = 100 x Cp x (64.6-32.6) + 1.10 x 10^5 + 4254.2912

Cp = 2.53 J/g.oC

So the specific heat for liquid MeOH would be 2.53 J/g.oC

Failed to expalin why for example he used the molar mass of Methanol(q2); and how the moles were cancelled out? I need something a more explicit answer.

Explanation / Answer

Molar mass of CH3OH = 12 + 4*1.008 + 16= 32.032 g/mol
Mass = 100 g
number of moles, n = mass /molar mass = 100/32.032 =3.122 mol
Let specific heat of liquid be C J/goC
Heat required to raise Temperature of liquid from 32.6oC to 64.6 oC,
Q1= m*C*(Tf-Ti)
     = 100*C*(64.6 - 32.6)
     =3200*C J
heat required to melt,
Q2 = n*Lv
       =3.122 mol*35.21 KJ/mol
       =109.926 KJ
        =109926 J
Heat required to raise Temperature of has from 64.6oC to 95.6 oC,
Q3= m*C*(Tf-Ti)
    = 100g*0.328cal/goC * (95.6 - 64.6) oC
   =1016.8 cal
   =1016.8 *4.184 J {since 1 cal = 4.184 J}
   =4254.3 J
Total heat reqired = Q1+Q2+Q3
122100 = 3200*C + 109926 + 4254.3
C=2.47 J/goC
Answer: 2.47 J/goC

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