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Experiment 14 Data and Calculations: Heat Effects and Calorimetry A. Specific Heat Trial Trial 2 Mass of stoppered test tube plus metal Mass of test tube and stopper Mass of calorimeter Mass of calorimeter and water Mass of water 139.18, 188-20 49.02. 65.10: 23.0c 139.18. 199.76. 60.52 65.6: 23.8 c Mass of metal Initial temperature of water in calorimeter Initial temperature of metal (assume 100°C unless directed to do otherwise) 100 °c 30. 2.c Equilibrium temperature of metal and water in calorimeter 106 c 30.6 cc 7.6 °C - 69.4 °c 160.00 Alumner final - Initial Afiquetal 94, 083 J/g°C _ J/g°C Specific heat of the metal (Eq. 3) Approximate molar mass of metal Unknown metal # B. Heat of Solution Mass of calorimeter plus water 188.20 29.04153 Mass of beaker (continued on following pe

Answer 1

Trial 1.

The heat lost by metal = heat gained by the water

i.e. -(m.C.$\Delta$T)_metal = (m.C.$\Delta$T)_water

Where m(metal) and m(water) are the masses of metal and water, respectively.

C(metal) and C(water) are the heat capacities of metal and water (4.184 J/g.^oC), respectively.

$\Delta$T(metal) and $\Delta$T(water) are the temperature changes of metal and water, respectively.

i.e. -65.101 g * C(metal) * (30.6-100) ^oC = 49.02 g * 4.184 J/g.^oC * (30.6-23) ^oC

i.e. C(metal) = 0.345 J/g.^oC

Trial 2.

-65.101 g * C(metal) * (30.2-100) ^oC = 60.58 g * 4.184 J/g.^oC * (30.2-23.8) ^oC

i.e. C(metal) = 0.357 J/g.^oC

Therefore, the average specific heat of the metal = (0.345+0.357)/2 = 0.351 J/g.^oC

Now, the