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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.
4 kwi 2024 · Given that 1 kJ is equal to 1000 joules, first convert the heat of vaporization to joules: 0.826 kJ/g * 1000 J/kJ = 826 J/g. Then, multiply this value by the mass of the ethanol: Q = 67.85 g * 826 J/g = 56066.1 J. Therefore, the heat energy required to boil 67.85 g of ethanol is 56066.1 joules.
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.
1. First, we need to convert the given mass of ethanol from grams to kilograms, as the heat of vaporization is given in kJ per gram. 37.55 g = 0.03755 kg. Step 2/3. 2. Next, we can use the formula: Heat energy = mass x heat of vaporization Heat energy = 0.03755 kg x 0.826 kJ/g Heat energy = 0.031 kJ.
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.
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
