# How much heat (in kJ) is required to raise the temperature of 202.8 g of ethanol...

How much heat (in kJ) is required to raise the temperature of 202.8 g of ethanol (mw=46.07) from 40.34 °C to its boiling point of  78.37 °C and then vaporize it completely creating ethanol gas at the boiling temperature?
(specific heat = 2.46 J/g°C, ΔHvap = 48.6 kJ/mol)

Given that

202.8 g of ethanol

Number of moles = 202.8 g of ethanol /46.07 g/ mole

= 4.40 moles

ΔHvap = number of moles * ΔHvap

= 4.40 mole *48.6 KJ/ mole

= 213.84 KJ

ΔH=m(c(Tb−T1)+ΔHvap)

= 202.9 g * 2.46 J/g°C (78.37 °C - 40.34 °C )+ 213.84 KJ

= 202.9 g * 2.46 J/g°C (38.03 °C )+ 213.84 KJ

= 18982.07 J +213.84 KJ

=18.982 kJ+213.84 KJ

=232.822 kJ

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