Question

Question #27d40

Answer 1

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)`