For apex it's 125.6 kJ on the 3.1.4 quiz 😊
1650kj
9460 kJ
To calculate the heat needed to vaporize sweat, you would need to know the specific heat of vaporization of sweat. Once you have that information, you can use the formula Q = mL, where Q is the heat needed, m is the mass of the substance (converted from volume using its density), and L is the specific heat of vaporization.
The heat of vaporization of ethanol is approximately 840 kJ/kg. To find the total heat required to vaporize 1.25 kg of ethanol, you can multiply the mass by the heat of vaporization: 1.25 kg * 840 kJ/kg = 1050 kJ.
Oh, dude, to vaporize 2kg of water at 100°C, you'd need about 2260 kilojoules of energy. It's like making a really intense cup of tea, but instead of sipping it, you're just turning it into steam. So, yeah, it's a pretty hot process, literally.
The heat of vaporization of gold is 158 kJ/kg. To find the total energy required to vaporize 2 kg of gold, you can use the equation: Energy = mass * heat of vaporization. Substitute the values to get: Energy = 2 kg * 158 kJ/kg = 316 kJ. Therefore, 316 kJ of energy is required to vaporize 2 kg of gold.
too much energy is needed to vaporize water
1oo calories for 1 g
To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10,700 kJ/kg. For 2 kg of aluminum, the total energy required would be 2 kg × 10,700 kJ/kg, equating to about 21,400 kJ. Therefore, around 21.4 million joules of energy is needed to vaporize 2 kg of aluminum.
1650kj
The amount of energy needed to vaporize 175 g of water depends on the temperature of the water. However, we shall assume it is 100 degrees C. We multiply 175 by 539 and get 94,325 calories. (Notice the small c). We could express it as 94 Calories if we were talking about the stuff on your dining room table.
To vaporize gold, we need to consider its molar enthalpy of vaporization. The molar enthalpy of vaporization of gold is approximately 334 kJ/mol. Since the molar mass of gold is about 197 g/mol, vaporizing 2 kg (2000 g) of gold requires: ( q = \frac{2000 , \text{g}}{197 , \text{g/mol}} \times 334 , \text{kJ/mol} \approx 3,385 , \text{kJ}. ) Thus, approximately 3,385 kJ of energy is required to vaporize 2 kg of gold.
It takes approximately 64,000 Joules of energy to melt 1kg of gold. Therefore, to melt 2kg of gold, you would need around 128,000 Joules of energy.
This vaporization energy is 18,19 kJ.
To calculate the energy required to vaporize 2 kg of aluminum, we use the heat of vaporization of aluminum, which is approximately 10,900 J/kg. Therefore, the energy required is 2 kg × 10,900 J/kg = 21,800 J, or 21.8 kJ. This is the amount of energy needed to convert 2 kg of aluminum from a liquid to a vapor at its boiling point.
9460 kJ
9460 kJ