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Specific Heat-- please help!
A piece of copper of mass 0.5 kg is heated to 610 degrees celcius and then dropped into a copper calorimeter of mass 0.11 kg containing 0.13 kg of water at 21 degrees celcius. If the specific heat of copper is 390 J/Kg/K, calculate the mass of water lost by boiling.
mC(Tf-Ti)= mC(Tf-Ti) + mc(Tf-Ti) + mLv
Qlost(copper)= Qgained (copper and water)
mC(tf-Ti) = mC(tf-ti) + mc(Tf-Ti) + mLv
0.5 x 390 x (610 -21) = 0.11 x 390 x (100-21) + 0.13 x 4200 x(100-21) + m2.25 x 10^6
I thought that, because the water is boiled and thus water is being lost (steam) then the final temp of the water and copper calorimeter is 100 degrees celcius.. am i on the right track? pls help me
Homework Statement
A piece of copper of mass 0.5 kg is heated to 610 degrees celcius and then dropped into a copper calorimeter of mass 0.11 kg containing 0.13 kg of water at 21 degrees celcius. If the specific heat of copper is 390 J/Kg/K, calculate the mass of water lost by boiling.
Homework Equations
mC(Tf-Ti)= mC(Tf-Ti) + mc(Tf-Ti) + mLv
The Attempt at a Solution
Qlost(copper)= Qgained (copper and water)
mC(tf-Ti) = mC(tf-ti) + mc(Tf-Ti) + mLv
0.5 x 390 x (610 -21) = 0.11 x 390 x (100-21) + 0.13 x 4200 x(100-21) + m2.25 x 10^6
I thought that, because the water is boiled and thus water is being lost (steam) then the final temp of the water and copper calorimeter is 100 degrees celcius.. am i on the right track? pls help me