physics calorimetry question?
(II) What will be the equilibrium temperature when a 265-g block of copper at 285°C is placed in a 145-g aluminum calorimeter cup containing 825 g of water at 12.0°C?
i got 10.6 degrees celsius is this correct? i know how to set up the equation but i am not sure on how to solve the algebra
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Hello Spongebob: Your answer of 10.6 degrees C cannot be correct because it is less than temperature of the constituents being thermally mixed.
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The First Law of Thermodynamics gives for the thermal mixing:
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Q = ( Delta UCU ) + ( Delta UAL ) + ( Delta UW ) + WB
There is no work done. ----> WB = 0.0
Assume perfectly insulated calorimeter. ----> Q ~ 0.0
Delta UCU = ( mCU ) ( CVCU ) ( TE - TCU1 )
Delta UCU = ( 265 ) ( 0.390 ) ( TE - 285 )
Delta UCU = (103.35 ) ( TE - 285 )
Delta UAL = ( mAL ) ( CVAL ) ( TE - TAL1 )
Delta UAL = ( 145 ) ( 0.900 ) ( TE - 12.0 )
Delta UAL = ( 130.5 ) ( TE - 12.0 )
Delta UW = ( mW ) ( CVW ) ( TE - TW1 )
Delta UW = ( 825 ) ( 4.185 ) ( TE - 12.0 )
Delta UW = ( 3452.625 )( TE - 12.0 )
0.0 = Delta UCU + Delta UAL + DeltauW
0.0 = ( 103.35 ) ( TE - 285 ) + ( 130.5 ) ( TE - 12.0 ) + ( 3452.625 ) ( TE - 12.0 )
0.0 = ( 103.35 + 130.5 + 3452.625 ) ( TE ) - 29455 - 1566 - 41432
(3686.475 ) ( TE ) = 72453
TE = 19.64 deg C <-----------------------------------
Now check the answer !!!
Delta UCU = ( 265 ) ( 0.390 ) ( 19.64 - 285 ) = - 27245 J
Delta UAL = ( 145 ) ( 0.900 ) ( 19.64 - 12.0 ) = 997 J
Delta UW = ( 825 ) ( 4.185 ) ( 19.64 - 12.0 ) = 26378 J
Delta UCU + Delta UAL + Delta UW = ( -27245 ) + 997 + 26378
SUM ( Delta U ) = 130 J ~ 0.0 [ OK !! }
Cp Copper = .385 kJ/kgC
Cp water = 4.186kJ/kgC
Cp Aluminum = .9kJ/kgC
For simplicity, convert 145g aluminum to .9/4.186 = 31.18g water
Here's the algebra:
Q (heat) = Cp (specific heat)ÃmassÃtemp change
Q lost = Q gained
.385kJ/kgC à .265kg à (285C-Tf) = 4.186kJ/kgCÃ(.825+.3118)kgÃ(Tf-12C)
29.077125-.102025Tf = 3.58396Tf - 43
72 = 3.685Tf
Tf = 19.53 C