Methanol boiling point = 64.6 °C
initial temperature = 0.8 °C
specific heat capacity = 2.53
Q = mcΔT = 156.g x 2.533 J/g.°C x (64.6 - 0.8)
Q = 25210 J = 252 x 102 J
now to boil, heat of vaporization = 35.21 kJ/mol
moles methanol = 156 g / 32.04 g/mol = 4.86891385768 mol
Q = 4.86891385768 mol x 35.21 kJ/mol = 171.434456929 kJ = 171434 J
Total heat required = 171434 J + 25180 J = 196614 J = 197 x 103 J
Answer: 197 x 103 J
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