Homework Answers
For this problem, we need to assume that the heat which is lost by the aluminum is equal to the heat gained by the water with no heat transfer outside the system.
The heat balancing equation is as follows -
−QAl = QH2O where Q = mcΔT where
Q = Heat Transferred,
m = mass of the substance,
c = specific heat of the substance (0.9 J/goC for Aluminium and 4.184 J/goC for water)
ΔT = change in temperature.
−mAl⋅cAl⋅(Tf−Ti) = mH2O⋅cH2O(Tf−Ti)
In ths , substituting given values, we get,
−(135)(0.9)(80−400) = m(4.184)(80−25)
38880 = m x 230.12
m = 38880 / 230.12
m = 168.96 g
Therefore, The mass of the water is 168.96 g
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a 135 g sample of aluminium at 400
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