Question 1

Three liquids are at temperatures of 10 ◦C, 22◦C, and 29◦C, respectively. Equal masses of the first two liquids are mixed, and the equi- librium temperature is 12◦C. Equal masses of the second and third are then mixed, and theequilibrium temperature is 25.9◦C.Find the equilibrium temperature when equal masses of the first and third are mixed.Answer in units of ◦C.

Question 2

$\begin{gathered} Liquid\text{ 1} \\ T_(L1)=10\text{ \degree C} \\ Liquid\text{ 2} \\ T_(L2)=22\text{ \degree C} \\ Liquid\text{ } \\ T_(L3)=29\text{ \degree C} \\ For\text{ Liquid 1 and Liquid 2} \\ T_1=12\text{ \degree C} \\ Q_(L1)=Q_(L2) \\ mC_(L1)\Delta T=mC_(L2)\Delta T \\ mC_(L1)(12\text{ \degree C-10\degree C})=mC_(L2)(22\text{ \degree-12\degree C}) \\ C_(L1)(2\text{ \degree C})=C_(L2)(10\text{ \degree C}) \\ C_(L1)=\frac{C_(L2)(10\text{ \degree C})}{2\text{ \degree C}} \\ C_(L1)=5C_(L2) \\ \\ For\text{ L}\imaginaryI\text{qu}\imaginaryI\text{d 2 and L}\imaginaryI\text{qu}\imaginaryI\text{d 3} \\ T_2=25.9\text{ \degree C} \\ Q_(L2)=Q_(L3) \\ mC_(L2)\Delta T=mC_(L3)\Delta T \\ mC_(L2)(25.9\text{ \degree C-22 \degree C})=mC_(L3)(29\text{ \degree C-25.9\degree C}) \\ C_(L2)(3.9\operatorname{\degree}\text{C})=C_(L3)(3\text{.1}\operatorname{\degree}\text{C}) \\ C_(L3)=\frac{C_(L2)(3.9\operatorname{\degree}\text{C})}{3\text{.1}\operatorname{\degree}\text{C}} \\ C_(L3)=(39C_(L2))/(31) \\ \\ For\text{ L}\imaginaryI\text{qu}\imaginaryI\text{d 1and L}\imaginaryI\text{qu}\imaginaryI\text{d 3} \\ T_3=? \\ mC_(L1)\Delta T=mC_(L3)\Delta T \\ mC_(L1)(T_3-10\text{ \degree C})=mC_(L3)(29\text{ \degree C-T}_3) \\ C_(L1)(T_3-10\operatorname{\degree}\text{C})=C_(L3)(29\operatorname{\degree}\text{C- T}_3) \\ But \\ C_(L1)=5C_(L2) \\ C_(L3)=(39C_(L2))/(31) \\ Hence \\ 5C_(L2)(T_3-10\operatorname{\degree}C)=(39C_(L2))/(31)(29\operatorname{\degree}C-T_3) \\ 5(T_3-10\operatorname{\degree}C)=(39)/(31)(29\operatorname{\degree}C-T_3) \\ 5T_3-50\text{ \degree C=}(1131)/(11)\text{ \degree C-}(39)/(31)T_3 \\ 5T_3+(39)/(31)T_3=(1131)/(11)\text{ \degree C+50 \degree C} \\ (194)/(31)T_3=(1681)/(11)\text{ \degree C} \\ \\ T_3=24.4\text{ \degree C} \\ \text{The equilibrium temperature is 24.4\degree C} \end{gathered}$

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