Question #27d40

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Answer 1 · 2016-08-03T13:33:57

To solve the problem the specific heat of methanol is needed. From internet data source we get it as #s_"methanol"=2.533"kJ"/(kg^@C)#

Let heat of vaporization of methanol be #L_"methanol"#

Heat required to raise the temperature of mehanol from #34^@C " to " 65^@C# is
#="mass"xx"sp.heat"xx"temperature rise"#

#=0.098kgxx 2.533"kJ"/(kg^@C)xx(65-34)^@C#

#~~7.7kJ#

Heat required to vaporise methanol at #34^@C# is
#="mass"xxL_"methanol"#
#=0.098xxL_"methanol"#

Now by the problem

#7.7+0.098xxL_"methanol"=93.487#

#:.L_"methanol"=(93.487-7. 7)/0.098"kJ"/(kg)#

#= 875.4"kJ"/(kg)#