Homework Answers
Here, we're setting the heat values derived given the equation:
q=mCsΔT
equal to each other, as such,
qwater+q(Cu)=0
∴qwater=−q(Cu)
Before we start, I will assume the density of water is 1.00 g/mL.
207g*4.184Jg⋅°C⋅(Tf−18.5°C)=−[30.5g*0.385Jg⋅°C⋅(Tf−99.8°C)]
866.08 (Tf) -16022.62 = -11.74 (Tf)+1171.85
877082 (Tf) = 17194.27
Tf=19.58°C Ans.
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