A) The internal energy change in the warm water is the heat transferred to the cold water. Likewise, the internal energy change in the cold water is the heat transferred from the hot water.
For the 50 g of liquid water at 0°C,
\begin{array}{ll}\mathrm{dU}=\mathrm{Q}\\\mathrm{dU} \approx \mathrm{dH} \approx \mathrm{mC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right)\\\mathrm{dU} \approx(50 \mathrm{~g})\left(4.19 \frac{\mathrm{J}}{\mathrm{g} \,\mathrm{K}}\right)\left(\mathrm{T}_{\mathrm{f}}-273 \mathrm{~K}\right)\\\mathrm{Q}=209.5 \mathrm{~T}_{\mathrm{f}}-57193.5 \mathrm{~J}\end{array}
For the 100 g of liquid water at 50°C,
\begin{array}{ll}\mathrm{dU}=-\mathrm{Q} \\ \mathrm{dU} \approx \mathrm{dH} \approx \mathrm{mC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right) \\\mathrm{dU} \approx(100 \mathrm{~g})\left(4.19 \frac{\mathrm{J}}{\mathrm{g} \,\mathrm{K}}\right)\left(\mathrm{T}_{\mathrm{f}}-323 \mathrm{~K}\right) \\ \mathrm{Q}=-419 \mathrm{~T}_{\mathrm{f}}+135337 \mathrm{~J}\end{array}
Set the two equations equal to each other to solve for T_{f}
\begin{array}{ll}-419 \mathrm{~T}_{\mathrm{f}}+135337 \mathrm{~J}=209.5 \mathrm{~T}_{\mathrm{f}}-57193.5 \mathrm{~J}\\\mathrm{T}_{\mathrm{f}}=306.33 \mathrm{~K}=\bf 33.33^{\circ} \mathrm{C}\end{array}
B) We will approach this problem under the assumption that the ice will melt before the water reaches 0°C. If our answer is unrealistic we know that the energy needed to melt the ice is greater than the energy that the water will give. If we get an unrealistic answer, we will go back and set T_f as 0°C and solve for the mass of ice that remains unmelted. In short, we are assuming the final phase is entirely liquid water.
For the 100 g of liquid water at 50°C,
\begin{array}{ll}\mathrm{dU}=-\mathrm{Q}\\\mathrm{dU} \approx \mathrm{dH} \approx \mathrm{mC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right)\\\mathrm{dU} \approx(100 \mathrm{~g})\left(4.19 \frac{\mathrm{J}}{\mathrm{g} \,\mathrm{K}}\right)\left(\mathrm{T}_{\mathrm{f}}-323 \mathrm{~K}\right)\\\mathrm{Q}=-419 \mathrm{~T}_{\mathrm{f}}+135337 \mathrm{~J}\end{array}
For the 50 g of ice at 0°C (the ice melts, then begins to increase in temperature),
\begin{array}{ll}\mathrm{dU}=\mathrm{Q}\\\mathrm{dU} \approx \mathrm{dH} \approx \mathrm{mC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right)+\Delta \underline{\mathrm{H}}_{\mathrm{fusion}}\\\mathrm{dU}=(50 \mathrm{~g})\left(4.19 \frac{\mathrm{J}}{\mathrm{g} \,\mathrm{K}}\right)\left(\mathrm{T}_{\mathrm{f}}-273 \mathrm{~K}\right)+333.55 \frac{\mathrm{J}}{\mathrm{g}}(50 \mathrm{~g})\\\mathrm{Q}=-40516 \mathrm{~J}+209.5 \mathrm{~T}_{\mathrm{f}}\end{array}
Set the two equations equal to each other to solve for T_{f}
\begin{array}{ll}-40516 \mathrm{~J}+209.5 \mathrm{~T}_{\mathrm{f}}=-419 \mathrm{~T}_{\mathrm{f}}+135337 \mathrm{~J}\\T_f=279.80\,K=\bf 6.80^{\circ}C\end{array}
C) There is no driving force for change, so there remains 50 grams of ice and 50 grams of liquid water
D) For this problem, it is difficult to assume anything specific about the final phase of water; both liquid and solid phases will be present in some quantity. Therefore, we will calculate the amount of heat that would be given off by the water as it cools to 0°C, and the amount of heat needed to heat the ice to 0°C.
For the water at 20°C
\begin{array}{ll} \mathrm{dU}=-\mathrm{Q} \\ \mathrm{dU} \approx \mathrm{dH} \approx \mathrm{mC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right) \\ \mathrm{dU} \approx(50 \mathrm{~g})\left(4.19 \frac{\mathrm{J}}{\mathrm{g} \,\mathrm{K}}\right)(273 \mathrm{~K}-293 \mathrm{~K}) \\\mathrm{Q}=-4190 \mathrm{~J}\end{array}
For the ice at –20°C
\begin{array}{ll} \mathrm{dU}=\mathrm{Q} \\ \mathrm{dU} \approx \mathrm{dH} \approx \mathrm{mC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right) \\ \mathrm{dU} \approx(50 \mathrm{~g})\left(2.11 \frac{\mathrm{J}}{\mathrm{g} \,\mathrm{K}}\right)(273 \mathrm{~K}-253 \mathrm{~K}) \\ \mathrm{Q}=2110 \mathrm{~J}\end{array}
Since these two Q values are not equal, there must be a phase change occurring, in addition to the temperature changes. For the 50 g of liquid water, Q = –4190 J. Thus for the 50 g that were originally ice, Q=4190 J. This is equal to the change in internal energy of the 50 g of ice:
\begin{array}{ll} \Delta \mathrm{U}_{\text {cooling }}+\Delta \mathrm{U}_{\text {melting }}=4190 \mathrm{~J}.\\\Delta \mathrm{U}_{\text {melting }}=4190 \mathrm{~J}-2110 \mathrm{~J}\\\left(333.55 \frac{\mathrm{J}}{\mathrm{g}}\right) \times\rm mass =2080 \mathrm{~J}\\\operatorname{mass}=6.24 \mathrm{~g}\end{array}
So 6.24 g of ice melts. This means the original 50g of liquid water plus 6.24g is the total mass of liquid, and the remainder is ice:
50g – 6.24g = 43.76 g ice
50g + 6.24g = 56.24 g liquid