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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.
25 sie 2023 · Explanation: To calculate the heat energy required to boil 66.45 g of ethanol, we use the given value of the heat of vaporization, which is 0.826 kJ/g. The equation for heat energy is q = mass x heat of vaporization.
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.
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.
Heat of vaporization= 0.826 KJg-1. Mass of Ethanol = 68.15 g. Let suppose temperature of ethanol is already at boiling point .
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
Convert the heat of vaporization from kilojoules to joules by multiplying 0.826 by 1000. Step 1 To calculate the heat energy required to boil ethanol, we need to use the formula:
