Hi lb_gilbert.
I'm not a chemistry expert, but I did find a similar question with a solution attached:
Two 20.0-g ice cubes at –17.0 °C are placed into 255 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
Answer
∆T for ice = (0 - (-17)) = +17 ºC
∆T for the water from the ice = (T - 0) - T ºC
∆T for the initial water only = (T - 25)
Heat quantity Q change for the ice = mol(ice)*cp(ice)*17 + enthalpy of fusion*mol(ice) + mol(ice)*cp(water)*T
Heat quantity Q change for the water = m0l(water)*cp(water)*(T - 25)
molar mass H2O = 18g/mol so 40g ice = 2.22 mol(ice) and 255 g water = 15.3 mol(water)
2.22*37.7*17 + 2.22*6010 + 2.22*75.3*T + 15.3*75.3*(T - 25) = 0
14740 + 167.2*T + 1152*T - 28802 = 0
1389*T = 14062
T = 10.1 ºC
Looking forward to more questions and your response!
