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3 years ago

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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 (c

water = 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

1 answer:

Alex787 [66]3 years ago

3 0

Answer:

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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