Final answer:
To boil the water, approximately 1.31 x 10^6 J of heat is required. The work done by the vapor pushing on the piston is approximately 0.021 atm*m³. The change in internal energy of the water is approximately 1.31 x 10^6 J.
Step-by-step explanation:
To calculate the amount of heat required to boil the water, we first need to calculate the mass of the water. Since the molecular weight of water is 18 g/mol, we can divide the given mass of 107 grams by the molecular weight to get the number of moles. So, 107 g / 18 g/mol = 5.94 mol.
Next, we need to calculate the heat required to vaporize the water. The latent heat of vaporization is given as 2260 J/g. So, the heat required is 5.94 mol * 2260 J/g * 107 g/mol = 1.31 x 10^6 J.
To calculate the work done by the vapor, we can use the ideal gas law. The volume of the vapor is given as Vf. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Since the pressure is constant and equal to 1 atm, we can rewrite the equation as V = (nRT)/P. Plugging in the values, V = (5.94 mol * 8.314 J/mol*K * 373.15 K) / 101325 Pa = 0.021 m³.
To calculate the work done by the vapor, we can use the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. Plugging in the values, W = 1 atm * 0.021 m³ = 0.021 atm*m³.
The change in internal energy of the water can be calculated using the first law of thermodynamics, which states that ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added, and W is the work done. Plugging in the values, ΔU = 1.31 x 10^6 J - 0.021 atm*m³ = 1.31 x 10^6 J - 2.13 J ≈ 1.31 x 10^6 J.