What is the mass of alcohol taken at a temperature of 28C if it is necessary to spend 12 kJ

What is the mass of alcohol taken at a temperature of 28C if it is necessary to spend 12 kJ of heat to heat it to the boiling point?

Problem data: t0 (initial alcohol temperature) = 28 ºС; Q (heat expended to heat alcohol to boiling) = 12 kJ (12 * 10 ^ 3 J).

Reference values: Cc (specific heat of alcohol) = 2500 J / (kg * K); tboil (alcohol boiling point) = 78 ºС.

The mass of alcohol taken can be expressed from the formula: Q = Cc * m * (tboil – t0), whence m = Q / (Cc * (tboil – t0)).

Let’s calculate: m = 12 * 10 ^ 3 / (2500 * (78 – 28)) = 0.096 kg (96 g).

Answer: 96 grams of alcohol was heated to a boil.