A 8.9-kg cube of copper (ccu 386 J/kg-K) has a temperature of 750 K. It is dropp

Question

A 8.9-kg cube of copper (ccu 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.5 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? K Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 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 x 106 J/kg) will be left after the water stops boiling? kg Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 3) Let's try this again, but this time add just the minimum amount of water needed to lower the temperature of the copper to 373 K. In other words, we start with the cube of copper at 750 K and we only add enough water at 293 K so that it completely evaporates by the time the copper reaches 373 K. Assume the resulting water vapor remaining at 373 K. How much water do we need? g Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question

Explanation / Answer

1)

it is in equilibrium:

-Q_copper = Q_water

-(m_Cu_*cCu_*deltaT) = m_water*_cwater*deltaT

-((8.9)(386)(x - 750) = (5.5)*(4186)*(x - 293)

-(3435.4x - 2576550) = 23023x - 6745739

-3435.4x + 2576550 = 23023x - 6745739

9322289 = 26458.4x

x = 352.34 K

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