lf = 3.35 x 105 J kg-1
This much amount of heat required to convert 1 kg of ice to liquid
Mani.Ra
It is the additional amount of heat [measured in (the "big") calories ] required to cause boiling/vaporization of a liquid already at the boiling/vaporization temperature. Heat of vaporization is the amount of heat (540 calories/gram) needed to change a specific amount of liquid (1 gram) at it's boiling point into a gas without changing it's temperature. -Qwasas
heat of fusion is 333.55 (kJ/kg) rephrasing the question to the metric system: How much heat will be required to change 2.27 Kg of ice at 0 degrees into water at 27 degrees? heat of fusion is 333 * 2.27 = 756 kjoules or 756000 joules (melts the ice) specific heat of water (liquid): cp = 4.185 J/gK (this is given in grams, so we have to convert) or cp = 4185 J/kgK multiplying 4185 * 2.27 * 27 = 256,500 joules total = 256,500 + 756000 = 1012500 joules or 241,800 callories or 960 BTU
The answer will depend on the temperature of the liquid water and the pressure.
When you heat the base of a liquid, the particles heat up and rise, as they become less dense. When they reach the top of the container, they cool and sink, as they now become more dense again. When they get to the bottom, the whole process starts again.
molar masse acetone: 58.08 g/mol 43.9g/58.08g/mol =0.75585mol the energy required for vaporization to a gas is... 0.75585molx29.1KJ/mol =21.995KJ
The heat required to convert ice at 0°C to water at 0°C is known as the latent heat of fusion. For water, this value is 334 J/g. Therefore, to convert 0.3 g of ice to water at the same temperature, the heat required is 0.3 g * 334 J/g = 100.2 Joules.
The heat required to convert 20 g of liquid water at 100°C to steam at 100°C is the heat of vaporization of water. This is 2260 J/g. Therefore, the total heat required is 20 g * 2260 J/g = 45200 J. This amount of heat is absorbed by the water as it changes phase from liquid to steam.
To convert ice at 0°C to liquid water at 0°C, 334 J/g of heat is needed (heat of fusion). To raise the temperature of liquid water from 0°C to 55°C, 4.18 J/g°C is required (specific heat capacity of water). The total heat required would be (mass of ice x 334 J) + (mass of ice x 4.18 J/g°C x temperature difference).
The measurement of how much heat energy is required for a substance to melt is called the heat of fusion. It is the amount of energy required to change a substance from a solid to a liquid at its melting point.
No heat (energy) is required to freeze water (from liquid to solid). Freezing RELEASES energy (heat), as it is an exothermic event. If you want to know how much energy is release, you need to know the heat of fusion for water, and then multiply that by the mass of water being frozen.
The energy required to boil a liquid is dependant on what the liquid is, as every liquid has it's own constant known as a specific heat capacity. This essentially boils down (ha ha) to the strength of the bonds between the molecules, which varies depending on the molecule itself. Water for instance has much stronger intermolecular attractions that, say, liquid nitrogen.
7.9
question makes no sense.....
The amount of heat that must be removed from steam to change it to a liquid is the latent heat of vaporization of water, which is approximately 2260 kJ/kg at atmospheric pressure. This is the amount of energy required for water to change from a gas to a liquid state at a constant temperature.
Can you help
The energy required to change ice to liquid water is known as the heat of fusion. For ice to water, the heat of fusion is 334 J/g. Therefore, to change 24.7 g of ice to liquid water, the energy required would be 24.7 g * 334 J/g = 8259.8 J.
The energy required to melt a solid into a liquid is known as the heat of fusion. This energy is needed to overcome the intermolecular forces holding the solid together and break down its crystal structure, allowing the particles to move freely in the liquid state. The amount of heat of fusion varies depending on the substance.