The question is: An insulated beaker with negligible mass contains liquid water with a mass of 0.225 kg and a temperature of 75.8 C. How much ice at a temperature of -23.6 C must be dropped into the water so that the final temperature of the system will be 31.0 C? Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*k, and the heat of fusion for water to be 334 KJ/kg. mice=? This is what I have so far: mw= 0.225 kg Tw= 75.8 C Ti= -23.6 C mi= ? Tf= 31.0 C Cw= 4190 J/kg(K) Ci= 2100 J/kg(K) Lf= 334 KJ/kg = 334000J/kg (I would have to convert this right?) With the information given I come up with these: Qice warms= mi ci (0- (-5)) Qice melts= mi Lf Qmix of waters= mi cw (31-0) Qwater cools= mw cw (31 - 75.8) Qwater cools = Qice warms + Qice melts + Q mix of waters so now we can solve the mass of ice. right? But when I calculate it the answer I is really weird.