m(water) = 159.0 g
T(water) = 22.8 oC
C(water) = 4.184 J/goC
m(alCOPPERloy) = 487.0 g
T(alCOPPERloy) = 89.5 oC
C(alCOPPERloy) = 0.387 J/goC
T = to be calculated
We will be using heat conservation equation
Let the final temperature be T oC
use:
heat lost by alCOPPERloy = heat gained by water and 3
m(alCOPPERloy)*C(alCOPPERloy)*(T(alCOPPERloy)-T) = m(water)*C(water)*(T-T(water)) + C3*(T-T(water))
487.0*0.387*(89.5-T) = 159.0*4.184*(T-22.8)+10.0*(T-22.8)
188.469*(89.5-T) = 675.256*(T-22.8)
16867.9755 - 188.469*T = 675.256*T - 15395.8368
T= 37.3543 oC
Answer: 37.35 oC
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