A 8.8-kg cube of copper (c_Cu = 386 J/kg-K) has a temperature of 750 K. It is dr

Question

A 8.8-kg cube of copper (c_Cu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.5 kg of water (c_water = 4186 J/kg-K) with an initial temperature of 293 K. What is the final temperature of the water-and-cube system? 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 Times 10^6 J/kg) will be left after the water stops boiling? 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?

Explanation / Answer

2). Since equilibrium and change in phase:

-m_cu*cCU*deltaT = m_evap*Latent Heat + m_water*cwater*deltaT

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

-(8.8*386*(373-1350) = m_evap*(2.26*10^6) + (5.5*4186*(373-293))
3318673 = m_evap*(2.26*10^6) + 1841840
m_evap = 0.6534 kg
This is the amount evaporated we need the amount remaining so subtract from the initial amount of water:
5.4 - 0.6534 = 4.74 kg remaining

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