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To calculate the heat energy required to boil 49.05 g of ethanol, you can use the given heat of vaporization and the mass of the ethanol. The calculation will utilize the formula Q = m × ΔHvap, where Q is the heat energy, m is the mass of the substance, and ΔHvap is the heat of vaporization.
12 gru 2020 · Calculate the total heat required (in joules) to convert 3.95 grams liquid ethanol (C2H5OH) at 25.0 °C to gas at 95.0 °C, given that the boiling point of ethanol is 78.5 °C, heat of vaporization is 40.5 kJ/mol, the specific heat of liquid ethanol is 2.45 J/g-K, and the specific heat of gaseous ethanol is 1.43 J/g-K
22 sie 2023 · To convert this to joules, we know that 1 kJ = 1000 joules. Therefore, the heat of vaporization for ethanol is 0.826 * 1000 = 826J/g. Now that we have converted the heat of vaporization into joules, we can multiply it by the mass of ethanol to calculate the heat energy required: 826J/g * 70.75g = 58419.5J.
To find the heat energy, we can use the formula: Heat energy (Q) = Mass (m) × Heat of vaporization (L) where Q is the heat energy, m is the mass of ethanol, and L is the heat of vaporization. Now, we can plug in the given values: Q = 54.65 g × 0.826 kJ/g Since we need the answer in joules, we need to convert kJ to J.
Question: The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 68.15 g of ethanol. heat energy: J. Show transcribed image text. There are 2 steps to solve this one.
See Answer. Question: The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 43.65 g of ethanol, heat energy. Show transcribed image text. Here’s the best way to solve it. 100% (1 rating) Share Share. Heat of vaporization = 0.826kJ/g it means at boiling point o … View the full answer.
Q: How much energy is needed to boil 45.0g of water at its boiling point? Water's heat of vaporization… A: Given: Mass of water = 45.0 g. And heat of vaporization of water = 40.65 KJ/mol.
