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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.
22 sie 2023 · The heat of vaporization for ethanol is given as 0.826 kJ/g. 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:
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.
17 sie 2023 · 1. 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.
30 sty 2023 · The Heat of Vaporization (also called the Enthalpy of Vaporization) is the heat required to induce this phase change. Figure \(\PageIndex{1}\): Heat imparts energy into the system to overcome the intermolecular interactions that hold the liquid together to generate vapor.
Heat of vaporization= 0.826 KJg-1. Mass of Ethanol = 68.15 g. Let suppose temperature of ethanol is already at boiling point .
