The amount of heat required to boil 111 grams of octane is 24.76 Joules.
The mass of Octane (C₂H₁₈) that is to be heated and boil is 111 Grams. The initial temperature of octane is 25.1 degrees Celsius,
The boiling point of octane is 125.6 degrees Celsius.
The specific heat of octane is 2.22 J/kg-K.
Converting the temperatures into Kelvin, the initial temperature is 298.25 K and final temperature is 398.75 K.
The heat required is given by,
Q = MCΔT
Cis specific heat, M is mass, and ΔT is the change in temperature, putting values.
Q = 0.111 x 2.22 x 100.5
Q = 24.76 Joules.
So, the heat required is 24.76 Joules.
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