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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.
17 sie 2023 · We know the heat of vaporization for ethanol is 0.826 kJ/g. This means that it takes 0.826 kJ of energy to vaporize 1 gram of ethanol. Step 2/5. 2. We need to find the heat energy required to boil 37.95 g of ethanol. To do this, we can use the following formula: Heat energy = Heat of vaporization x Mass of substance.
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.
Heat of vaporization= 0.826 KJg-1. Mass of Ethanol = 68.15 g. Let suppose temperature of ethanol is already at boiling point .
The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 20.55 g of ethanol. heat energy: J
The specifi c heat of water is 1.00 cal/(g · °C); the heat of fusion of water is 79.7 cal/g; and the heat of vaporization of water is 540 cal/g. a. How much energy (in calories) is needed to melt 45 g of ice at 0.0 °C and warm it to 55 °C?
