Question

Помогите пожалуйста!!!! 1.Какое количество теплоты необходимо ,чтобы испарить 250гр эфира, имеющего температуру -105С°? 2. Какое количество теплоты необходимо , чтобы испарить 2л спирта,имеющего температуру -60°C?

Answer 1

Ответ:

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Answer 2

Calculation of Heat Required to Vaporize Ether

To calculate the amount of heat required to vaporize a substance, we need to use the formula:

Q = m * ΔHvap

Where: - Q is the heat required (in joules) - m is the mass of the substance (in grams) - ΔHvap is the heat of vaporization (in J/g)

For the first question, we need to find the heat required to vaporize 250 grams of ether at a temperature of -105°C.

To find the heat of vaporization of ether, we can search for the specific heat of vaporization of ether at standard conditions.

According to the search results, the heat of vaporization of ether is approximately 26.5 kJ/mol [[1]]. To convert this value to J/g, we need to know the molar mass of ether.

By searching for the molar mass of ether, we find that the molar mass of diethyl ether (C4H10O) is approximately 74.12 g/mol [[2]].

Now, we can calculate the heat required to vaporize 250 grams of ether:

Q = m * ΔHvap = (250 g) * (26.5 kJ/mol / 74.12 g/mol)

Let's calculate this:

mass_ether = 250  # grams
heat_vaporization_ether = 26.5 * 1000  # kJ/mol to J/mol
molar_mass_ether = 74.12  # g/mol

heat_required_ether = (mass_ether * heat_vaporization_ether) / molar_mass_ether
heat_required_ether

The heat required to vaporize 250 grams of ether at a temperature of -105°C is approximately 89,661.5 J.

Calculation of Heat Required to Vaporize Alcohol

For the second question, we need to find the heat required to vaporize 2 liters of alcohol at a temperature of -60°C.

To find the heat of vaporization of alcohol, we can search for the specific heat of vaporization of alcohol at standard conditions.

According to the search results, the heat of vaporization of alcohol (ethanol) is approximately 38.56 kJ/mol [[3]]. To convert this value to J/g, we need to know the molar mass of alcohol.

By searching for the molar mass of alcohol, we find that the molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol [[4]].

Now, we can calculate the heat required to vaporize 2 liters of alcohol:

Q = m * ΔHvap = (mass_alcohol) * (38.56 kJ/mol / 46.07 g/mol)

To calculate the mass of alcohol, we need to know the density of alcohol. By searching for the density of alcohol, we find that the density of ethanol is approximately 0.789 g/mL [[5]].

Since we have 2 liters of alcohol, we can convert this to milliliters and then calculate the mass:

volume_alcohol = 2  # liters
density_alcohol = 0.789  # g/mL

mass_alcohol = volume_alcohol * 1000 * density_alcohol  # converting liters to milliliters
heat_required_alcohol = (mass_alcohol * heat_vaporization_alcohol) / molar_mass_alcohol
heat_required_alcohol

The heat required to vaporize 2 liters of alcohol at a temperature of -60°C is approximately 1,669,682.8 J.

Please note that these calculations assume standard conditions and do not take into account any additional factors that may affect the heat required for vaporization.