Final answer:
The heat added to convert 1 cm³ of water to 1000 cm³ of steam at atmospheric pressure is 2.2632 kJ. The work done by the water against the atmosphere is 0.16867 kJ, and the change in internal energy of the water is 2.09453 kJ.
Step-by-step explanation:
To calculate the heat added to a system where water changes to steam at atmospheric pressure, we can use the concept of latent heat of vaporization. Assuming that the mass of water is 1g (since the density of water is 1g/cm³), the heat needed to vaporize water at 100°C and 1 atm pressure is found using the molar heat of vaporization which is 40.7 kJ/mol for water.
Firstly, we convert the mass of water to moles (molar mass of water = 18.0 g/mol):
- Number of moles = mass (g) / molar mass (g/mol) = 1 g / 18.0 g/mol = 0.0556 mol
Then, we multiply the number of moles by the molar heat of vaporization to find the heat absorbed:
- Heat absorbed (Q) = number of moles * molar heat of vaporization = 0.0556 mol * 40.7 kJ/mol = 2.2632 kJ
The work done by the water against the atmosphere can be calculated using the formula:
Work done (W) = pressure * change in volume = 1.01 x 10µ N/m² * 1.67 x 10⁻³ m³ = 168.67 J or 0.16867 kJ (since 1 J = 1 N*m).
Using the first law of thermodynamics, the change in internal energy (ΔU) of the water is equal to the heat added minus the work done by the system:
- ΔU = Q - W
- ΔU = 2.2632 kJ - 0.16867 kJ = 2.09453 kJ