Question

При испарении спирта выделяется 300 Дж энергии.Сколько было спирта?

Answer 1

А там точно 300Дж, а не 300МДж?

900000Дж/кг-питома теплота пароутворення (l)

Q=lm

m=Q÷l

m=300Дж÷900000Дж/кг=0,000333кг

Answer 2

To determine the amount of alcohol (ethanol) that evaporated and released 300 Joules of energy, we need some additional information. Specifically, we need to know the mass of the alcohol that evaporated and the heat of vaporization of ethanol.

The heat of vaporization (also known as enthalpy of vaporization) is the amount of energy required to convert one unit mass of a liquid into a gas at a constant temperature and pressure. For ethanol, the typical value for the heat of vaporization is approximately 841 kJ/kg (kilojoules per kilogram) at its boiling point, which is about 78.37°C.

Now, let's assume that the heat of vaporization of ethanol is 841,000 J/kg.

We can use the formula:

Energy released = Mass of alcohol evaporated * Heat of vaporization

Given Energy released = 300 J and Heat of vaporization = 841,000 J/kg, we can rearrange the formula to solve for the mass of alcohol evaporated:

Mass of alcohol evaporated = Energy released / Heat of vaporization

Mass of alcohol evaporated = 300 J / 841,000 J/kg

Mass of alcohol evaporated ≈ 0.000356 kg (approximately 0.356 grams)

So, approximately 0.356 grams of alcohol (ethanol) evaporated to release 300 Joules of energy.