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Calculate the heat released when ethanol is cooled from 400K to 100K. The boiling point of ethanol is 351 K and the melting point is 159K. The heat capacity of solid , liquid , gaseous ethanol are 111.46J (mol•K) , 112.4 J/(mol•K) , and 87.53 J/(mol•K)... the enthalpies for ethanol are Hfus= 4.9 kj/mol and the Hvap= 38.56 kj/mol.

Respuesta :

The total heat of a process is the sum of all the heat involved in the process. So, the total heat is the sum of all sensible and latent heat in the whole process. For this case, the flow of the release of heat is,

sensible heat from 400 K to the boiling point (351 K) ---> latent heat due to condensation ------> sensible heat from 351 K to melting point (159 K) -----> latent heat due to freezing --------> sensible heat from 159 K to 100 K

Total heat released =  87.53 J/(mol•K) (400 K - 351 K) + 38560 J/mol + 112.4 J/(mol•K) ( 351 K - 159 K ) + 4900 J/mol + 111.46J / (mol•K) ( 159 K - 100 K) = 75905.91 J / mol

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