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3 December, 22:43

A 155.0 g piece of copper at 168 oc is dropped into 250.0 g of water at 20.9 oc. (the specific heat of copper is 0.385 j/goc.) calculate the final temperature of the mixture. (assume no heat loss to the surroundings.)

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  1. 3 December, 22:55
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    Note that

    The specific heat of water is 4.184 J / (g - C)

    Let x = the final equilibrium temperature.

    Heat loss by the copper is

    (155 g) * (0.385 J / (g - C)) * (168 - x C) = 59.675 (168 - x) J

    Heat gain by the water is

    (250 g) * (4.184 J / (g-C)) * (x - 20.9 C) = 1046 (x - 20.9) J

    Because there is no heat loss to the surroundings, therefore

    59.675 (168 - x) = 1046 (x - 20.9)

    168 - x = 17.5283 (x - 20.9)

    18.5283x = 534.3415

    x = 28.84 °C

    Answer: 28.8 °°C nearest tenth)
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