b kJ kJ 10 42 dynamite ton 1 6 1908 10 4 19 10 4 ton dynamite The

B kj kj 10 42 dynamite ton 1 6 1908 10 4 19 10 4 ton

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(b) 8.014 × 10 10 kJ × kJ 10 4.2 dynamite ton 1 6 × = 1.908 × 10 4 = 1.9 × 10 4 ton dynamite The energy released by this rainstorm is equivalent to explosion of 19,000 tons of dynamite. 5.97 Find the heat capacity of 1.7 × 10 3 gal H 2 O. C - g 1 J 184 . 4 cm 1 g 1 L 1 cm 10 1 qt 1.057 L 1 gal 1 qt 4 O H gal 10 7 . 1 C o 3 3 3 2 3 O H 2 × × × × × × × = = 2.692 × 10 7 J/ ° C = 2.7 × 10 4 kJ/ ° C; then, bricks 18,000 or 10 8 . 1 kg 1.8 brick 1 g 10 1 kg 1 J 0.85 C - g 1 C 1 J 10 692 . 2 4 3 o o 7 × = × × × × × Check . (1.7 × ~16 × 10 6 )/(~1.6 × 10 3 ) 17 × 10 3 bricks; the units are correct. 5.98 (a) J 10 27 . 3 9 . 3274 C) 4 . 100 C 1 . 30 ( Cu g 0 . 121 K g J 385 . 0 q 3 o o Cu - × = = × × = The negative sign indicates the 3.27 × 10 3 J are lost by the Cu block. (b) J 10 1 . 3 3138 C) 1 . 25 C 1 . 30 ( O H g 0 . 150 K g J 184 . 4 q 3 o o 2 O H - 2 × = = × × = The positive sign indicates that 3.14 × 10 3 J are gained by the H 2 O.
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5 Thermochemistry Solutions to Exercises 127 (c) The difference in the heat lost by the Cu and the heat gained by the water is 3.275 × 10 3 J – 3.138 × 10 3 J = 0.137 × 10 3 J = 1 × 10 2 J. The temperature change of the calorimeter is 5.0 ° C. The heat capacity of the calorimeter in J/K is 0.137 × 10 3 J/K. 10 3 4 . 27 C 0 . 5 1 J o × = = × Since O H 2 q is known to one decimal place, the difference has one decimal place and the result has 1 sig fig. If the rounded results from (a) and (b) are used, J/K. 10 4 C 0 . 5 J 10 2 . 0 C o 3 r calorimete × = × = (d) T) ( g 0 . 150 K g J 184 . 4 J 10 275 . 3 q - 3 O H 2 Δ × × = × = Δ T = 5.22 ° C; T f = 25.1 ° C + 5.22 ° C = 30.3 ° C 5.99 (a) From the mass of benzoic acid that produces a certain temperature change, we can calculate the heat capacity of the calorimeter. C kJ/ 78 . 3 7755 . 3 acid benzoic g 1 kJ 38 . 26 observed change C 642 . 1 acid benzoic g 235 . 0 o o = = × Now we can use this experimentally determined heat capacity with the data for caffeine. caffeine kJ/mol 10 22 . 4 caffeine mol 1 caffeine g 2 . 194 C 1 kJ 7755 . 3 caffeine g 265 . 0 rise C 525 . 1 3 o o × = × × (b) The overall uncertainty is approximately equal to the sum of the uncertainties due to each effect. The uncertainty in the mass measurement is 0.001/0.235 or 0.001/0.265, about 1 part in 235 or 1 part in 265. The uncertainty in the temperature measurements is 0.002/1.642 or 0.002/1.525, about 1 part in 820 or 1 part in 760. Thus the uncertainty in heat of combustion from each measurement is kJ 6 760 4220 kJ; 5 820 4220 kJ; 16 265 4220 kJ; 18 235 4220 = = = = The sum of these uncertainties is 45 kJ. In fact, the overall uncertainty is less than this because independent errors in measurement do tend to partially cancel. 5.100 Plan . Use the heat capacity of H 2 O to calculate the energy required to heat the water. Use the enthalpy of combustion of CH 4 to calculate the amount of CH 4 needed to provide this amount of energy, assuming 100% transfer. Assume H 2 O(l) is the product of combustion at standard conditions. Solve. required kJ 272 J 10 720 . 2 C) 0 . 25 C 0 . 90 ( C - g J 184 . 4 kg 1 g 1000 O H kg 00 . 1 5 o o o 2 = × = × × × CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ) g ( O H ) g ( CH H O(l) H H 2 ) g ( CO H H 2 f 4 f 2 f 2 f rxn o o o o o Δ Δ Δ + Δ = Δ
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5 Thermochemistry Solutions to Exercises 128 = –393.5 kJ + 2(–285.83 kJ) – (–74.8 kJ) – (0) = –890.4 kJ 4 4 4 4 2 CH g 90 . 4 899 . 4 CH mol CH g 04 . 16 kJ 4 . 890 CH mol 1 kJ 10 720 . 2 = = × × × 5.101 (a) Mg(s) + 2H 2 O(l) Mg(OH) 2 (s) + H 2 (g) ) s ( Mg H ) l ( O H H 2 (g) H H ) s ( Mg(OH) H H f 2 f 2 f 2 f rxn o o
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