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.
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
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.
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)
That's called heat of fusion. Going from a liquid to a gas is heat of vaporization.
Thisa is enthalpy of fusion and is different for each substance.
It is the latent heat of melting (liquefaction).
less than 1 degrees Celsius
50 times of steam
water has a high heat of vapourization.it absorbs much heat as it changes from liquid to gas.it has the capacity of absorbing heat with minimum of change in its own temperature
The specific heat capacity of water does not change much within-phase (ie, as a solid it has one specific heat capacity, as a liquid/gas it has another)
because the specific heat of water is very high
To change the temperature of water from 27ºC to 32ºC will depend on the mass of water that is present. Obviously, the more water, the more heat it will take. This can be calculated as follows:q = heat = mC∆T where m is the mass of water; C is sp. heat = 4.184 J/g/deg and ∆T is 5ºC (change in temp).
50 times of steam
55 Ml.
Heat required to have such a change of state is called latent heat. If L J/kg is the latent heat per kg of water then for M kg of water we need M* L joule of heat energy
water has a high heat of vapourization.it absorbs much heat as it changes from liquid to gas.it has the capacity of absorbing heat with minimum of change in its own temperature
The specific heat capacity of water does not change much within-phase (ie, as a solid it has one specific heat capacity, as a liquid/gas it has another)
Water has a high specific heat capacity (relative to metals and other conductors), making it a poor conductor of heat (takes too much energy to change the temperature).
because the specific heat of water is very high
To change the temperature of water from 27ºC to 32ºC will depend on the mass of water that is present. Obviously, the more water, the more heat it will take. This can be calculated as follows:q = heat = mC∆T where m is the mass of water; C is sp. heat = 4.184 J/g/deg and ∆T is 5ºC (change in temp).
The equation is q = mC∆T where q is the heat; m is the mass of water; C is the specific heat of water (1 cal/g/deg); and ∆T is the change in temperature.
Heat stored is based on a change in temperature. If you heat the water to 75 C, then you can extract that heat when the water is cooled to 25 C again. 1 cubic meter of water is 1,000,000 grams Heat absorbed or released = specific heat capacity x mass x change in temperature. = (1 cal / g C) x (1,000,000 g) x (75 - 25)C = 50,000,000 calories lower temperature change means less heat stored.
If there is too much water, people are flooded out of their homes. If there is not enough water, you have drought, and that leads to heat stroke, thirst, and hunger.
This heat is: 5 x 4,18 x m = 20,9 x m (in calories) where m is the mass of water.