Question

Three liquids are at temperatures of 7 ◦C,
20◦C, and 34◦C, respectively. Equal masses of
the first two liquids are mixed, and the equilibrium temperature is 11◦C. Equal masses of
the second and third are then mixed, and the
equilibrium temperature is 22.6◦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 )

$C_1$ - specific heat of the first third 

$C_2$ - specific heat of the second  liquid

$C_3$ - specific heat of the third liquid

Temperature of liquids: $T_1=7 ◦C,T_2=20◦C, T_3=34◦C$

Temperature of 1+2 liquids mix: $T_{12}=11^oC$

Temperature of 2+3 liquids mix: $T_{23}=22.6^oC$

Temperature of 1+3 liquids mix: $T_{13} - ?$

When the first two are mixed:

$m C₁ (T₁ − T_{12}) + m C₂ (T₂ − T_{12}) = 0 \\ C₁ (7− 11) + C₂ (20 − 11) = 0\\ 4C_1=9C_2\\ C_1=2.25C_2$

When the second and therd are mixed:

$m C_2 (T_2 − T_{23}) + m C_3 (T_3 − T_{23}) = 0 \\ C_2 (20−22.6) + C₂ (34 −22.6) = 0\\ 2.6C_2=11.4C_3\\ C_2=4.38C_3$

When the first and therd are mixed:

$m C_1 (T_1 − T_{13}) + m C_3 (T_3 − T_{13}) = 0 \\ C_1 (7−T_{13}) + C_3 (34 −T_{13}) = 0\\ C_1=2.25C_2=2.25(4.38C_3)=9.86C_3\\ 9.86C_3 (7−T_{13})=-C_3(34 −T_{13})\\ 9.86 (7−T_{13})=-(34 −T_{13})\\ T_{13}=9.5^oC$

$T_{13}=9.5^oC$