Question

Three liquids are at temperatures of 11 ◦C,

Answer 1

Answer:

  18.6 °C

Explanation:

Let k1 represent the specific heat of the first liquid relative to that of the 2nd. We can find the value of k1 from the first mixture.

  (11 °C)(k1) +(19 °C)(1.0) = (14 °C)(1 +k1)

  (19 -14) = (14 -11)k1

  k1 = 5/3

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Let k3 represent the specific heat of the third liquid relative to that of the 2nd. We can find the value of k3 from the second mixing that was done.

  (19 °C)(1) +(29 °C)(k3) = (24.5 °C)(1 +k3)

  (29 -24.5)k3 = (24.5 -19)

  k3 = 5.5/4.5 = 11/9

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The equilibrium temperature (T) resulting from mixing the first and third liquids will be ...

  (11 °C)(5/3) +(29 °C)(11/9) = T(5/3 +11/9)

  165/9 +319/9 = T(26/9)

  T = (165 +319)/26 = 18 8/13 ≈ 18.6 . . . . degrees C

The equilibrium temperature when equal masses of the first and third liquids are mixed will be about 18.6 °C.