Question

A 603.0 g/hr stream of liquid methyl alcohol, also called methanol, (CH3OH) at 8.10 atm and 26.0°C was held at constant pressure, vaporized and brought to 210.0°C. At what rate must heat be supplied to this system? Assume that methyl alcohol vapor behaves ideally for the temperature range and pressure given.

Answer 1

Answer:

$Q=\frac{1385270 J}{hr}$

Explanation:

Methanol's molecular weight: $M=32 g/mol$

Methanol's heat of vaporization: $\Delta H_{vap}=69.69 kJ/mol$

Ideal gas heat capacity (Cp):

$Cp=\frac{5}{2}*R$ where R is the gas constant

$Cp=\frac{5}{2}*8.314 \frac{J}{mol*C}=20.78\frac{J}{mol*C}$

The heat needed to vaporize and bring the gs to 210°C is:

$Q=\frac{603 g/hr}{32 g/mol}*(69690\frac{J}{mol}+20.78\frac{J}{mol*C}*(210-26)C)$

$Q=\frac{1385270 J}{hr}$