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

Sagot :

[tex]\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}=\frac{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}=\frac{39C_{L2}}{31} \\ Hence \\ 5C_{L2}(T_3-10\operatorname{\degree}C)=\frac{39C_{L2}}{31}(29\operatorname{\degree}C-T_3) \\ 5(T_3-10\operatorname{\degree}C)=\frac{39}{31}(29\operatorname{\degree}C-T_3) \\ 5T_3-50\text{ \degree C=}\frac{1131}{11}\text{ \degree C-}\frac{39}{31}T_3 \\ 5T_3+\frac{39}{31}T_3=\frac{1131}{11}\text{ \degree C+50 \degree C} \\ \frac{194}{31}T_3=\frac{1681}{11}\text{ \degree C} \\ \\ T_3=24.4\text{ \degree C} \\ \text{The equilibrium temperature is 24.4\degree C} \end{gathered}[/tex]

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