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    Equilibrium state of a mixture.

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    Question:

    Three liquids are at temperatures of 10 degrees C, 20 degrees C and 30 degrees C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 17 degrees C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 28 degrees C. Find the equilibrium temperature when equal masses of the first and third are mixed.

    I am not sure how to approach this problem. All my other problems in this chapter included much more specific information. The following equations are used in this chapter of Energy in Thermal Processes.

    Q = m c deltaT where m = mass, c = specific heat, and delta T = change in temperature.

    Q = +/- mL where L = latent heat.

    Is there a pattern I'm looking for?

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    Solution Preview

    Please see the attached file.
    However you might want to try to solve it using these tips:

    There is no phase transition here, so there is no need to care about latent heat.
    The common factor in the two processes is liquid 2. In the first mix liquid 1 gains heat that liquid 2 ...

    Solution Summary

    The equilibrium state of a mixture is provided. Equal masses of temperatures are given.

    $2.49

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