Temperature change of iron = -40.2 oC
Expanation
volume of water = 10.5 mL
mass of water = (volume of water) * (density of water)
mass of water = (10.5 mL) * (1.00 g/mL)
mass of water = 10.5 g
Process 1 : All water is heated from 30.0 oC to 100 oC
H1 = (mass of water) * (specific heat water) * (final temperature - initial temperature)
H1 = (10.5 g) * (4.184 /g.oC) * (100 oC - 30.0 oC)
H1 = 3075.24 J
Process 2 : All water is converted to steam
H2 = (mass of water) * (heat of vaporization)
H2 = (10.5) * (2256.42 J/g)
H2 = 23692.4 J
Total heat gained by water = H1 + H2
Total heat gained by water = 3075.24 J + 23692.4 J
Total heat gained by water = 26767.6 J
Total heat gained by water = 26.77 kJ
Total heat lost by iron = -Total heat gained by water
Total heat lost by iron = -26.77 kJ
Temperature change of iron = (Total heat lost by iron) / [(mass of iron) * (specific heat iron)]
Temperature change of iron = (-26.77 kJ) / [(1.50 kg) * (0.444 kJ/kg.oC)]
Temperature change of iron = -40.2 oC
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