Answer:
a mass of water required is mw= 1273.26 gr = 1.27376 Kg
Explanation:
Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:
Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L
where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation
therefore
mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )
replacing values
mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg
