A 8.5-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.4 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 × 106 J/kg) will be left after the water stops boiling?
1). Since its in equilibrium: -Q_copper = Q_water -(m_Cu_*cCu_*deltaT) = m_water*_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 Temp 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 -(8.5*386*(373-1350) = m_evap*(2.26*10^6) + (5.4*4186*(373-293)) 3205537 = m_evap*(2.26*10^6) + 1808352 1397185 = m_evap*(2.26*10^6) m_evap = .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
