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5 Thermochemistry Solutions to Exercises 5.57 Analyze/Plan. Follow the logic in Sample Exercise 5.7. Solve. 9bomb = = 30.57°C 23.44°C = 7.13°C At constant volume, qv = E and are very similar. ≈ = = -56.0kJ = -25.454 = C₆H₄O₂ = kJ/mol C₆H₄O₂ 5.58 (a) C₆H₅OH(s) (b) 11.66 At constant volume, qv = and AH are very similar. ≈ = = = = 3.054 kJ 1 mol C₆H₅OH C₆H₅OH 5.59 Analyze. Given: specific heat and mass of glucose, for calorimeter. Find: heat capacity, C, of calorimeter. Plan. All heat from the combustion raises the temperature of the calorimeter. Calculate heat from combustion of glucose, divide by for calorimeter to get kJ/°C. = 24.72°C 20.94°C = 3.78°C Solve. (a) = 3.500 glucose 15.57 glucose 1 = = (b) Qualitatively, assuming the same exact initial conditions in the calorimeter, twice as much glucose produces twice as much heat, which raises the calorimeter temperature by twice as many °C. Quantitatively, =7.56°C Check. Units are correct. T is twice as large as in part (a). The result has 3 sig figs, because the heat capacity of the calorimeter is known to 3 sig figs. 5.60 (a) 1 (b) °C 1.440 1 = 30.046 = sample 120