Given the following data regarding octane:
Heat capacity: 255.6 J/(g °C)
Boiling point: 127°C
Heat of vaporization: 41.4 kJ/mole
AIT: 220°C
Density: 0.70 g/mL
Calculate the heat (Q) required to heat 15.0 mL of octane from room temperature (25°C) to the boiling point. Report the value in kJ

Question

Quizplus · Accepted Answer

The heat required to heat the octane from room temperature to the boiling point is:Q = m * Cp * ΔT + m * Lwhere:m is the mass of octaneCp is the heat capacity of octaneΔT is the change in temperatureL is the heat of vaporizationThe mass of octane is:m = V * d = 15.0 mL * 0.70 g/mL = 10.5 gThe change in temperature is:ΔT = Tb - Tr = 127°C - 25°C = 102°CThe heat required to heat the octane from room temperature to the boiling point is:Q = 10.5 g * 255.6 J/(g °C) * 102°C + 10.5 g * 41.4 kJ/mole = 274 kJTherefore, the correct answer is A.

Given the Following Data Regarding Octane