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Standard vaporization enthalpy of benzene at its boiling point is `30.8 kJ mol^(-1)`, for how long would a `100W` electric heater have to operate in order to vaporize a `100g` sample of benzene at its boiling temperature? |
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Answer» Standard vaporisation enthalpy of benzene `=30.8 kJ mol^(-1)` Power of electric heater `=100 W` Molar mass of `C_(6)H_(6)=72+6=78 am u` Heat required to vaporise `78 g C_(6)H_(6)=30.8xx1000J` heat required to vaporise `100 g C_(6)H_(6)` `=(30.8xx1000xx100)/78` `=39900 J` `100 W=100 J s^(-1)` `100 J` heat is given in `1 s` `39900` heat is given in `=(39900xx1)/(100)=399 s=6.6 min` |
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