How many kilojoules of heat are released when 2500 grams of gaseous ethanol at its boiling point are converted to liquid ethanol and then cooled down by 55 degrees celsius?

The molar heat of vaporization of ethanol =39.3kj/mol;
specific head of liquid ethanol= 2.26j/gxdegreesC

Please Please help I dont understand these type of problems of how to do the unit conversions properly :(

1 Answer
Feb 7, 2018

#DeltaH_"process"^@~=-2500*kJ#

Explanation:

I was not going to touch this problem, but then I saw that you have included the relevant thermodynamic parameters, so bravo...

We interrogate TWO thermodynamics processes....

#(i)# #"the enthalpy of condensation of GASEOUS EtOH"#

And #(ii)# #"the enthalpy released by liquid EtOH as it cools from"#
#"its boiling point to"# #55# #""^@C#

For #(i)# we gots....

#EtOH(g) rarr EtOH(l) + 39.3*kJ#

#"Moles of EtOH"=(2500*g)/(46.07*g*mol^-1)=54.27*mol#

And thus energy released for #(i)#...#=54.27*molxx-39.3*kJ*mol^-1=-2132.6*kJ#..and the negative sign means that energy is released to the surroundings as the process takes place.

And for #(ii)#, we gots....#DeltaT_"process"=T_f-T_i=-55*K#

#DeltaH_"process"^@=(-55*K)xx2500*gxx2.26*J*K^-1*g^-1#

#=-310.75*kJ#...

And as is typical the enthalpy associated with the phase change, dwarfs the energy involved in the single phase....

#DeltaH_"total"^@=(i)+(ii)=-(2132.6+310.75)*kJ=-2443.4*kJ#.