One kg of ice at 0°C is added to one kg of boiling water. The mixture comes to equilibrium. What is the change in entropy in cal/K of the system? (Lf = 80cal/g) a. 300 b. 100 c. 200 d. 50 e. 25 Qcold = -Qhot Mw*Cw*dT + Lf*Mw = -Mw*Cw*dT + Mw*Lv All has Mw in common, therefore it should cancel out Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv (1cal/g)(Tf - 0) + 80cal/g = -(1cal/g)(Tf-100) + 540cal/g; I converted the latent heat of vaporation (2.26x10^6J/kg)*(1/4186cal*kg/J*g) ~ 540cal/g after equating the two sides, I got 280C for Tf My answer for Tf doesn't seem reasonable because the Final Temp should be in between 0-100C Anyways, My other problem is getting the dQ I have no idea how to get it...or maybe I do..when it doubt I just ask people out! dS = dQ/T if my T is right then just finding dQ problem Thanks in advance folks.