Hmmm...
How about this, does this seem correct?
mm = 81.9g
mw = 100mL
Tim = 92°C
Tf = 29°C
Tiw = 25°C
Tfw = 29°C
cw = 4.19 J/g·°C (this is the number the school operates with)
cm = ?
81.9g(cm)(29°C - 92°C)+100g(4.19J/g·°C)(29°C - 25°C) = 0...
This is what I've got:
Iron mass: 81.9g
Initial iron temp.: 92'C
Water volume: 100mL
Initial water temp.: 25'C
Final water temp.: 29'C
What the the experimental heat capacity of iron?
Thanks :)
Bjorn