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Unformatted text preview: CHAPTER 6 THERMOCHEMISTRY: ENERGY FLOW AND CHEMICAL CHANGE 6.1 The sign of the energy transfer is defined from the perspective of the system. Entering the system is positive, and leaving the system is negative. 6.2 No , an increase in temperature means that heat has been transferred to the surroundings, which makes q positive. 6.3 E = q + w = w, since q = 0. Thus, the change in work equals the change in internal energy. 6.4 Plan: Remember that an increase in internal energy is a result of the system (body) gaining heat or having work done on it and a decrease in internal energy is a result of the system (body) losing heat or doing work. Solution: The internal energy of the body is the sum of the cellular and molecular activities occurring from skin level inward. The bodys internal energy can be increased by adding food, which adds energy to the body through the breaking of bonds in the food. The bodys internal energy can also be increased through addition of work and heat, like the rubbing of another persons warm hands on the bodys cold hands. The body can lose energy if it performs work, like pushing a lawnmower, and can lose energy by losing heat to a cold room. 6.5 a) electric heater b) sound amplifier c) light bulb d) automobile alternator e) battery (voltaic) 6.6 Plan: Use the Law of Conservation of Energy. Solution: The amount of the change in internal energy in the two cases is the same. By the law of energy conservation, the change in energy of the universe is zero. This requires that the change in energy of the system (heater or air conditioner) equals an opposite change in energy of the surroundings (room air). Since both systems consume the same amount of electrical energy, the change in energy of the heater equals that of the air conditioner. 6.7 Heat energy; sound energy (impact) Kinetic energy (falling text) Potential energy (raised text) Mechanical energy (raising of text) Chemical energy (biological process to move muscles) 6.8 The change in a systems energy is E = q + w . If the system receives heat, then its q final is greater than q initial so q is positive. Since the system performs work, its w final < w initial so w is negative. E = q + w E = (+425 J) + (425 J) = 0 J . 6.9 q + w = 255 cal + (428 cal) = 683 cal 6.10 Plan: A system that releases thermal energy has a negative value for q and a system that has work done on it has a positive value for work. Solution: E = -675 J + (530 cal x 4.184 J/cal) = 1542.52 = 1.54 x 10 3 J 6-1 6.11 E = q + w = ( ) 3 10 J 0.615 kJ 1 kJ + ( ) 3 10 cal 4.184 J 0.247 kcal 1 kcal 1 cal = 1648.4 = 1.65 x 10 3 J 6.12 Plan: Convert 6.6 x 10 10 J to the other units using conversion factors....
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- Spring '10