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14 February, 05:53

A 8.6-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.1 kg of water (cwater = 4186 J/kg-K) with an initial temperature of 293 K. 1) What is the final temperature of the water-and-cube system?2) If the temperature of the copper was instead 1350 K, it would cause the water to boil. How much liquid water (latent heat of vaporization = 2.26

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  1. 14 February, 08:04
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    1. Since its in equilibrium:

    Qcopper = Qwater

    (mCu*cCu*deltaT) = (mwater*cwater*deltaT)

    ((8.5) (386) (x-750)) = (5.4) * (4186) * (x-293)

    (3281x - 2460750) = 22604.4x - 6623089

    3281x + 2460750 = 22604.4x - 6623089

    9083839.2 = 26185.4x

    x = 346.904K = Final Temperature

    2. Since equilibrium and change in phase:

    mcu*cCU*deltaT = mevap*Latent Heat + mwater*cwater*deltaT

    Final Temperature is 373K because that is when liquid water becomes gas water

    (8.5*386 * (373-1350) = mevap * (2.26*10^6) + (5.4*4186 * (373-293))

    3205537 = mevap * (2.26*10^6) + 1808352

    1397185 = mevap * (2.26*10^6)

    mevap =.61822 kg

    This is the amount evaporated we need the amount remaining so subtract from the initial amount of water:

    5.4 -.61822 = 4.782 kg remaining
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