Let M be the mass of steam condensed.Amount of heat lost by the steam due to latent heat of condensation and due to cooling to 80^{\\circ}C =M\ imes 540+m(100-80) Amount of heat absorbed by the water and calorimeter system= 1100\ imes 1\ imes (80-15)+20\ imes 1 \ imes (80-15) From conservation of heat energy, =M\ imes 540+M(100-80) =1100\ imes 1\ imes (80-15)+20\ imes 1 \ imes (80-15) \\implies M=130g=0.130kg