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kherson [118]
2 years ago
12

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.
Chemistry
1 answer:
mafiozo [28]2 years ago
4 0

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}

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