Question

Ethanol boils at 78.4 °C with \DeltaΔHvap = 38.6 kJ/mol. A 0.200-mol sample of ethanol is heated from some colder temperature up to 78.4 °C, which requires 1.05 kJ of heat, and then vaporized. What will be the total amount of heat required (for both the heating and the vaporizing)?

Answer 1

Answer:

$Q_T=8.77kJ$

Explanation:

Hello,

In this case, the total heat is computed via:

$Q_T=n*\Delta H_{vap}+1.05 kJ$

It means, by adding all the involved heat, which simply result:

$Q_T=0.200mol*38.6 kJ/mol+1.05 kJ\\Q_T=8.77kJ$

Thus, such heat must be provided to the ethanol, that is why it remains positive.

Best regards.