How many calories of energy are needed to vaporize a 20.2 g sample of liquid ethanol at its normal boiling point of 78 oC?
q = mass ethanol x delta Hvap ethanol
To find out how many calories of energy are needed to vaporize a sample of liquid ethanol, we need to use the formula:
Q = m * ΔHv
where:
Q is the amount of heat energy required (in calories)
m is the mass of the substance (in grams)
ΔHv is the heat of vaporization of the substance (in calories/gram)
First, we need to determine the heat of vaporization of ethanol. The heat of vaporization is the amount of energy required to convert one gram of a substance from the liquid phase to the gas phase at a specific temperature.
The heat of vaporization of ethanol at its normal boiling point (78 °C) is approximately 204.4 calories/gram.
Given that the mass of the sample is 20.2 grams and the heat of vaporization is 204.4 calories/gram, we can now calculate the amount of heat energy required:
Q = m * ΔHv
Q = 20.2 g * 204.4 cal/g
Calculating this gives us:
Q = 4,134.88 calories
Therefore, approximately 4,134.88 calories of energy are needed to vaporize a 20.2 g sample of liquid ethanol at its normal boiling point of 78 °C.