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1 July, 06:45

009

Three liquids are at temperatures of 14 ◦C, 25◦C, and 33◦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 16◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 27.9 ◦C. Find the equilibrium temperature when equal masses of the first and third are mixed.

Answer in units of ◦C.

+1
Answers (1)
  1. 1 July, 07:13
    0
    16.1°C

    Explanation:

    When the first two are mixed:

    m C₁ (T₁ - T) + m C₂ (T₂ - T) = 0

    C₁ (T₁ - T) + C₂ (T₂ - T) = 0

    C₁ (14 - 16) + C₂ (25 - 16) = 0

    -2 C₁ + 9 C₂ = 0

    C₁ = 4.5 C₂

    When the second and third are mixed:

    m C₂ (T₂ - T) + m C₃ (T₃ - T) = 0

    C₂ (T₂ - T) + C₃ (T₃ - T) = 0

    C₂ (25 - 27.9) + C₃ (33 - 27.9) = 0

    -2.9 C₂ + 5.1 C₃ = 0

    C₂ = 1.76 C₃

    Substituting:

    C₁ = 4.5 (1.76 C₃)

    C₁ = 7.91 C₃

    When the first and third are mixed:

    m C₁ (T₁ - T) + m C₃ (T₃ - T) = 0

    C₁ (T₁ - T) + C₃ (T₃ - T) = 0

    (7.91 C₃) (14 - T) + C₃ (33 - T) = 0

    (7.91) (14 - T) + 33 - T = 0

    111 - 7.91T + 33 - T = 0

    8.91T = 144

    T = 16.1°C
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