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The heat of fusion for water is 6.02 kJ/mol. So we need to convert 1 gram of water into moles of water: 1 g H2O x 1 mol H2O/18.02 g H2O = 0.055 moles H2O Now we can find the heat released from freezing those 0.055 moles: 0.055 moles H2O x 6.02 kJ/1 mol H2O = 0.334 kJ of heat released Of course if we use the proper number of significant figures (since you only wrote 1g) it would be 0.3 kJ of heat released.

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15y ago
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14y ago

However much you but into the water. Or, however much difference there is between the water and its environment. For example, if you have a cup of water of 25*C and the ambient temperature is 15*C, the water give out 10*C of energy. I'll let you work out the actual number in J

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11y ago

The heat of evaporation of 1 gram of water at 100C at atmospheric pressure is 538.5 calories.

1 gram of water condensing from a 100% quality of saturation (dry steam) at 100C and atmospheric pressure releases 538.5 calories.

For lower quality the enthalpy is lower. Such that the latent heat of evaporation is reduced by the amount of moisture in the steam.

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14y ago

Enthalpy of fusion is the energy it takes to change from a solid to a liquid state (melt ice at 0°C to water at 0°C)

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14y ago

That's called heat of fusion. Going from a liquid to a gas is heat of vaporization.

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8y ago

Thisa is enthalpy of fusion and is different for each substance.

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8y ago

It is the latent heat of melting (liquefaction).

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13y ago

less than 1 degrees Celsius

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Q: How much heat requried to change 1gm of ice into water?
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