Question #112138
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.
1
Expert's answer
2020-04-27T10:15:25-0400

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

When the first two are mixed:


mC1(T1T12)+mC2(T2T12)=0mC_1(T_1−T _{ 12} ​ )+mC_2(T_2−T _{ 12} ​ )=0

C1(1215)+C2(2515)=0C2=0.3C1C_1(12−15 ​ )+C_2(25−15 ​ )=0\to C_2=0.3C_1

When the second and therd are mixed:


mC3(T3T32)+mC2(T2T32)=0mC_3(T_3−T _{ 32} ​ )+mC_2(T_2−T _{ 32} ​ )=0

C3(3830.5)+C2(2530.5)=0C2=1.36C3C_3(38−30.5 ​ )+C_2(25−30.5 ​ )=0\to C_2=1.36C_3

When the first and therd are mixed:


mC1(T1T13)+mC3(T3T13)=0mC_1(T_1−T _{ 13} ​ )+mC_3(T_3−T _{ 13} ​ )=0

10.3(15T13)+11.36(38T13)=0\frac{1}{0.3}(15-T_{13})+\frac{1}{1.36}(38-T_{13})=0

T13=19.2°CT_{13}=19.2\degree C


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