## Essential University Physics: Volume 1 (3rd Edition)

We first find the energy it would take to get either to 0 degrees celsius: $Q_{water}=mc_{water}\Delta T =(1)(4.184)(5)=20.92 \ kJ$ $Q_ice = mc_{ice}\Delta T =(1)(2.05)(40)=82 \ kJ$ Now, we see how much energy is involved in turning water into ice: $Q=334m$ Thus, we see that all of the mixture will be 0 degrees Celsius, and $\frac{82-20.92}{334}=.18 \ kg$ of it will become ice. In the end, everything will be at 0 degrees Celsius, and .18 kilograms will be ice, while the other .82 kilograms will be water.