Answer:
[tex]Q=\frac{1385270 J}{hr}[/tex]
Explanation:
Methanol's molecular weight: [tex]M=32 g/mol[/tex]
Methanol's heat of vaporization: [tex]\Delta H_{vap}=69.69 kJ/mol[/tex]
Ideal gas heat capacity (Cp):
[tex]Cp=\frac{5}{2}*R[/tex] where R is the gas constant
[tex]Cp=\frac{5}{2}*8.314 \frac{J}{mol*C}=20.78\frac{J}{mol*C}[/tex]
The heat needed to vaporize and bring the gs to 210°C is:
[tex]Q=\frac{603 g/hr}{32 g/mol}*(69690\frac{J}{mol}+20.78\frac{J}{mol*C}*(210-26)C)[/tex]
[tex]Q=\frac{1385270 J}{hr}[/tex]
