So in order to figure out what case I'm in do I calculate the energy it takes to bring water from 20C to 0C, so
(50)(75.3)(0-20) = -75,300J
Then figure out the amount of energy it takes to bring the ice from -20C to 0, so
(20)(38.07)(0- (-20)) = 15,228
-75,300+15,228 = 60,072 left...
1. You have 0.9 L of water in an insulated (thermally isolated) cup at 20 C. You add 360 g of ice at −20 C in the cup to cool the water. Describe the final state of your system in terms of (i) number of moles of ice (ii) number of moles of water and (iii) the final temperature.