1) The amount of heat needed to change water into vapour can be obtained as
"\\Delta Q = Lm = 2.260\\cdot10^6\\,\\mathrm{J\/kg}\\cdot3\\cdot10^{-3}\\,\\mathrm{kg} = 6780\\,\\mathrm{J}."
2) The work done by water is "A=P\\Delta V = 1.01\\cdot10^5\\,\\text{Pa}\\cdot2.45\\cdot10^{-3}\\,\\mathrm{m^{-3}} = 247.45\\,\\mathrm{J}."
3) The change of internal will be "\\Delta U = \\Delta Q - A = 6780\\,\\mathrm{J} - 247.45\\,\\mathrm{J} \\approx 6532.6\\,\\mathrm{J}."
Comments
Leave a comment