what volume and mass of steam at 100 degrees C and 1.00 atm would, during condensation, release the same amount of heat as 100.cm^3 of liquid water during freezing? What do you note, qualitatively, about the relative volumes and masses of steam and liquid water required to release the same amount of heat?
To find the volume and mass of steam that would release the same amount of heat as 100 cm^3 of liquid water during freezing, we need to consider the heat of fusion for water and the heat of condensation for steam.
First, let's find the amount of heat released during the freezing of 100 cm^3 of liquid water. The heat of fusion for water is 334 J/g. However, we need to convert the volume of water to mass by considering its density. The density of water is approximately 1 g/cm^3. Thus, the mass of 100 cm^3 of water can be calculated as:
Mass = Volume x Density
Mass = 100 cm^3 x 1 g/cm^3
Mass = 100 g
Now, the amount of heat released during the freezing of 100 g of water can be calculated as:
Heat released = Mass x Heat of Fusion
Heat released = 100 g x 334 J/g
Heat released = 33,400 J
To find the volume and mass of steam that would release the same amount of heat, we need to consider the heat of condensation for steam. The heat of condensation for water is also 334 J/g. However, in this case, we need to account for the fact that steam is gaseous, so we'll need to use the ideal gas law to calculate the volume and mass.
The ideal gas law states: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Given that the pressure is 1.00 atm and the temperature is 100 degrees C, which is 373 K, we can rearrange the ideal gas law to solve for V:
V = nRT/P
To find n, the number of moles, we need to convert the mass of steam to moles using the molar mass of water. The molar mass of water is approximately 18 g/mol.
n = Mass/Molar mass
n = Mass/18
Substituting this value into the ideal gas law equation:
V = (Mass/18) x R x T/P
Now, we can substitute the given values and solve for V:
V = (Mass/18) x 0.0821 L.atm/mol.K x 373 K / 1.00 atm
Simplifying the equation and converting to cm^3:
V = (Mass/18) x 0.0821 x 373 / 1,000
Now, we can set up the equation to find the mass of steam that releases the same amount of heat as 100 cm^3 of liquid water during freezing:
Heat released by steam = Mass x Heat of Condensation
33,400 J = Mass x 334 J/g
Now, we can compare the mass and volume of steam required to release the same amount of heat as 100 cm^3 of liquid water. It is important to note that the absolute values of the mass and volume will depend on the specific conditions, such as pressure and temperature. However, qualitatively, we can infer that the mass of steam required will be much less than the mass of liquid water due to the high heat of condensation of steam compared to the heat of fusion of water. Additionally, the volume of steam required will likely be much greater than the volume of liquid water due to the lower density of the gas state compared to the liquid state.