La Casa Verde

CHAPTER 8 ENERGY, POWER AND CLIMATE CHANGE
Exercise 8.1 (page 194)
1. 2. 3. 4. 5. 6. 7. D A C D D B Approximately 15% lost in furnace, 40% lost in heat exchanger, 10% lost as friction in turbine and the generator, 35% output as electrical energy. Therefore the power station is about 35% efficient. Chemical to thermal and light Chemical to thermal and kinetic Sound to electrical Chemical tothermal and kinetic and sound Electrical to light and thermal Electrical to light Electrical to thermal Electrical to sound Thermal to electrical Nuclear to thermal, sound and light

8.

These are suggested and selected answers only. The Exercise numbers refer to the 2009 revised edition. Owners of earlier editions should use page numbers to identify each Exercise. © IBID Press 2009

CHAPTER 8ENERGY, POWER AND CLIMATE CHANGE
Exercise 8.3 (page 208)
1. 2. 3. 4. B C D It is determined in joules per gram J g –1 or kilojoules per gram kJ g –1 as bomb calorimetry is used to determine the value and this technique requires only small masses of a sample. A mole can be a large mass. And we buy fuels by mass/volume not moles thus it is more useful to use this unit. To increase the surface areaof the coal to allow for a greater rate of combustion. Energy required to convert the water to steam = mcΔT + mLV Since there is 65% moisture content, there is 65 g of water per 100g of coal The heat energy absorbed to turn 65 grams of water into steam would be: Q = 65g × 4.18 Jg –1 K –1 × (100 oC – 20 oC) + 65g × (22.5 × 10 2 J g –1) = 167986 J = 168 kJ The energy density in the other 35 gramsof lignite is 28 kJ g –1 × 35 = 980 kJ The total usable energy in 100 grams = 980 – 168 kJ = 812kJ For one gram this would be 8.1 kJ g –1 The energy density as it is mined will be 8.1 kJ per gram less than when the coal is dried. = 28 kJ g –1 – 8.1 kJ g –1 The energy density of the coal as it is mined is 19.9 kJ g –1 . 7. (a) Assuming the hydrogen and oxygen is converted to steam, the total amountof steam = 30% + 30% of the remaining 70% = 51% 51% of 1000 tonnes = 510 tonnes. (b) 510 × 24 × 7 = 8.6 x 104 tonnes (c) 1 tonne = 1000 kg 1dm3 = 1kg 8.6 × 104 tonnes × 1000 = 8.6 × 107 dm3

5. 6.

These are suggested and selected answers only. The Exercise numbers refer to the 2009 revised edition. Owners of earlier editions should use page numbers to identify each Exercise. © IBID Press 2009CHAPTER 8 ENERGY, POWER AND CLIMATE CHANGE
8. Crude oil is used in the petrochemical industry to produce many products such as plastics, polymers, pharmaceuticals, synthetic textiles and fabrics. Other fuels such as LPG and LNG have a higher energy density than petrol and there are more supplies of gases than crude oil. The petrol engine is only 25% efficient and a greater efficiency can beobtained from cars that run on liquid petroleum gas. LPG is cleaner than petrol as it burns more efficiently and it contains less pollutants. In the future, the petrochemical industries will need feedstock to continue to produce products for consumers. (a) (b) C5 H4 + 6 O2
→

9.

5CO2 + 2H2O

One mole of coal (64 grams per mole) requires 6 moles of oxygen (32 grams per mole).The mass ingrams in 1000 tonne of coal

= 1000 tonne × 1000 kg × 1000g = 109 g. The number of mole of coal = 109 g / 64 g per mol = 1.56 × 107 mol. The number of mol of oxygen = 1.56 × 107 mol × 6 = 9.36 × 107 mol. Therefore, the mass of oxygen required = 32 × 9.36 × 107 g = 300 x 107g = 3.0 ×106 kg = 3000 tonnes of oxygen. (c) (d) 10. Volume of oxygen = 25 dm3 × 9.36 × 107 mol = 2.34 × 109 dm3. Volume of air= 5 × 2.34 × 109 dm3 = 1.2 × 1010 dm3.

(a) Since 35% efficient heat must be supplied at 500 MW / 0.35 = 1429 MW (b) 1 kg consumed for 31.5 × 106 J s–1. So for 1429 × 106 J s–1 the kg s–1 is 1429 / 31.5 = 45.4 kg s–1. (c) The amount of heat entering the cooling towers = 1429 – 500 = 929 MW. Q = mcΔT. So Q / t = mcΔT / t. Therefore, m/t = Q / cΔT m / t = 929 × 106 / 4180 × 10 = 2.2 × 104 kg...