[tex]\begin{gathered} Liquid\text{ 1} \\ T_{L1}=10\text{ \degree C} \\ Liquid\text{ 2} \\ T_{L2}=20\text{ \degree C} \\ Liquid\text{ 3} \\ T_{L3}=38\text{ \degree C} \\ For\text{ Liquid 1 and 2} \\ T1=\text{ 12\degree C} \\ Q_{L1}=Q_{L2} \\ mC1(T1-T_{L1})=mC2(T_{L2}-T1) \\ Equal\text{ mass} \\ C1(T1-T_{L1})=C2(T_{L2}-T1) \\ Solving\text{ C1 in term of C2} \\ C1=\frac{C2(T_{L2}-T1)}{T1-T_{L1}} \\ C1=\frac{C2(20\text{ \degree C}-12\text{ \degree}C)}{12\text{ \degree C}-10\text{ \degree C}} \\ C1=4C2 \\ For\text{ Liquid 2 and 3} \\ T2=35.8\text{ \degree C} \\ Q2=Q3 \\ mC2(T2-T_{L2})=mC3(T_{L3}-T2) \\ Equal\text{ mass} \\ C2(T2-T_{L2})=C3(T_{L3}-T2) \\ Solving\text{ C3 in term of C2} \\ C3=\frac{C2(T2-T_{L2})}{T_{L3}-T2} \\ C3=\frac{C2(35.8\text{ \degree C}-20\text{ \degree C})}{38\text{ \degree C-35.8 \degree C}} \\ C3=7.2C2 \\ For\text{ Liquid 1 and 3} \\ T3=? \\ Q1=Q3 \\ mC1(T1-T_{L1})=mC3(T_{L3}-T2) \\ Equal\text{ mass} \\ C1(T3-T_{L1})=C3(T_{L3}-T3) \\ But \\ C1=4C2 \\ C3=7.2C2 \\ 4C2(T3-T_{L1})=7.2C2(T_{L3}-T3) \\ 4(T3-T_{L1})=7.2(T_{L3}-T3) \\ 4T3-4T_{L1}=7.2T_{L3}-7.2T3 \\ Solving\text{ T3} \\ 4T3+7.2T3=7.2T_{L3}+4T_{L1} \\ 11.2T3=7.2T_{L3}+4T_{L1} \\ 11.2T3=7.2(38\text{ \degree C})+4(10\text{ \degree C}) \\ 11.2T3=273.6\text{ \degree C}+40\text{\degree C} \\ 11.2T3=313.6\text{ \degree C} \\ T3=\frac{313.6\text{ \degree C}}{11.2} \\ T3=28\text{ \degree C} \\ The\text{ equilibrium temperature is 28\degree C} \end{gathered}[/tex]
