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7 April, 14:58

Three liquids are at temperatures of 9 ◦C, 24◦C, and 30◦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 equilibrium temperature is 26.2 ◦C.

Find the equilibrium temperature when equal masses of the first and third are mixed.

Answer in units of ◦C

+2
Answers (1)
  1. 7 April, 15:08
    0
    14.8°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 (9 - 15) + C₂ (24 - 15) = 0

    -6 C₁ + 9 C₂ = 0

    C₁ = 1.5 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 (24 - 26.2) + C₃ (30 - 26.2) = 0

    -2.2 C₂ + 3.8 C₃ = 0

    C₂ = 1.73 C₃

    Substituting:

    C₁ = 1.5 (1.73 C₃)

    C₁ = 2.59 C₃

    When the first and third are mixed:

    m C₁ (T₁ - T) + m C₃ (T₃ - T) = 0

    C₁ (T₁ - T) + C₃ (T₃ - T) = 0

    (2.59 C₃) (9 - T) + C₃ (30 - T) = 0

    (2.59) (9 - T) + 30 - T = 0

    23.3 - 2.59T + 30 - T = 0

    3.59T = 53.3

    T = 14.8°C
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