Answer to Question #112138 in Molecular Physics | Thermodynamics for Mariela

Question

Three liquids are at temperatures of 12 ◦C,
25◦C, and 38◦C, respectively. Equal masses of
the first two liquids are mixed, and the equilibrium temperature is 15◦C. Equal masses of
the second and third are then mixed, and the
the equilibrium temperature is 30.5
◦C.
Find the equilibrium temperature when
equal masses of the first and third are mixed.
Answer in units of ◦C.

Answer 1

m - mass of each liquids (all masses are equal ).

When the first two are mixed:

$$mC_1(T_1−T _{ 12} ​ )+mC_2(T_2−T _{ 12} ​ )=0$$

$$C_1(12−15 ​ )+C_2(25−15 ​ )=0\to C_2=0.3C_1$$

When the second and therd are mixed:

$$mC_3(T_3−T _{ 32} ​ )+mC_2(T_2−T _{ 32} ​ )=0$$

$$C_3(38−30.5 ​ )+C_2(25−30.5 ​ )=0\to C_2=1.36C_3$$

When the first and therd are mixed:

$$mC_1(T_1−T _{ 13} ​ )+mC_3(T_3−T _{ 13} ​ )=0$$

$$\frac{1}{0.3}(15-T_{13})+\frac{1}{1.36}(38-T_{13})=0$$

$$T_{13}=19.2\degree C$$