Question

Austin Community College Chemistry Specific Heat Report Form Name Date: METAL Unknown Collected Data: METAL: Trial 1 Trial 2 Trial 3 lo. Mass of empty calorimeter Mass of calorimeter plus water Mass of metal Initial temperature of water Final temperature of water no.10 2.0 70.813 o C 5.5 1.30 loo a Initial temperature of metal Final temperature of metal 129

please help :/

Answer 1

For Trial 1,

heat gained by water = mCpdT

                                  = 165.301 x 4.18 x 3 = 2073 J

heat lost by metal = 70.813 x Cp x 74 = 2073

specific heat of metal = 2073/(70.813 x 74) = 0.395 J/g.oC

Trial 2,

heat gained by water = mCpdT

                                  = 163.331 x 4.18 x 4 = 2731 J

heat lost by metal = 69.672 x Cp x 73 = 2731

specific heat of metal = 0.537 J/g.oC

Trial 3,

heat gained by water = mCpdT

                                  = 158.755 x 4.18 x 8 = 5309 J

heat lost by metal = 69.202 x Cp x 69 = 5309

specific heat of metal = 1.112 J/g.oC

Mean specific heat of metal = 0.681 J/g.oC

Standard deviation = sq.rt.[(sum(x-mean)^2/2]

                               = sq.rt.[(0.085 + 0.021 + 0.186)/2]

                               = 0.382

% standard devaition = 38.2%