Question

Three liquids are at temperatures of 2°C, 21°C, and 34°C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 5 C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 30.1°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_1C 1 - specific heat of the first third C_2C 2 - specific heat of the second liquid

C_3C 3- specific heat of the third liquid

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

1 =7◦C,T 2 =20◦C,T 3=34◦C

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

12 =11 oC

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

23=22.6 oC

Temperature of 1+3 liquids mix: T_{13} - ?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_2mC₁(T₁−T 12)+mC₂(T₂−T 12 )=0C₁(7−11)+C₂(20−11)=04C \1=9C 2C 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_3mC 2 (T 2 −T 23 )+mC 3 (T 3 −T 23 )=0C 2(20−22.6)+C₂(34−22.6)=02.6C 2 =11.4C 3C2 =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^oCmC 1(T1 −T13)+mC 3(T3−T13 )=0C1 (7−T13 )+C 3(34−T 13 )=0C1 =2.25C2=2.25(4.38C3 )=9.86C39.86C3 (7−T13 )=−C3(34−T13 )9.86(7−T13)=−(34−T13 )T13 =9.5 oCT_{13}=9.5^oCT 13 =9.5 o C: