Chapter 6

# Chapter 6 - CHAPTER 6 THERMOCHEMISTRY: ENERGY FLOW AND...

This preview shows pages 1–3. Sign up to view the full content.

6-1 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 The internal energy of the body is the sum of the cellular and molecular activities occurring from skin level inward. The body’s internal energy can be increased by adding food, which adds energy to the body through the breaking of bonds in the food. The body’s internal energy can also be increased through addition of work and heat, like the rubbing of another person’s warm hands on the body’s 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 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 system’s 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. The change in energy is (+425 J) + (-425 J) = 0 J . 6.9 E = q + w = -255 cal + (-428 cal) = -683 cal 6.10 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. So, E = -675 J + (525 cal x 4.184 J/cal) = 1521.6 = 1.52 x 10 3 J 6.11 E = q + w = () 3 10 J 0.615 kJ 1kJ    + 3 10 cal 4.184 J 0.247 kcal 1 kcal 1 cal = 1648.4 = 1.65 x 10 3 J

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6-2 6.12 C( s ) + O 2 ( g ) CO 2 ( g ) + 3.3 x 10 10 J (1.0 ton) a) E (kJ) = (3.3 x 10 10 J) 3 1kJ 10 J    = 3.3 x 10 7 kJ b) E (kcal) = (3.3 x 10 10 J) 3 1 cal 1 kcal 4.184 J 10 cal = 7.887 x 10 6 = 7.9 x 10 6 kcal c) E (Btu) = (3.3 x 10 10 J) 1Btu 1055 J = 3.12796 x 10 7 = 3.1 x 10 7 Btu 6.13 CaCO 3 ( s ) + 9.0 x 10 6 kJ CaO( s ) + CO 2 ( g ) (5.0 ton) a) E (J) = (9.0 x 10 6 kJ) 3 10 J = 9.0 x 10 9 J b) E (cal) = (9.0 x 10 6 kJ) 3 10 J 1 cal 1 kJ 4.184 J = 2.15105 x 10 9 = 2.2 x 10 9 cal c) E (Btu) = (9.0 x 10 6 kJ) 3 10 J 1 Btu 1055J = 8.5308 x 10 6 = 8.5 x 10 6 Btu 6.14 E (J) = (4.1 x 10 3 Calorie) 3 10 cal 4.184 J 1 Calorie 1 cal = 1.7154 x 10 7 = 1.7 x 10 7 J E (J) = (4.1 x 10 3 Calorie) 3 3 10 cal 4.184 J 1 kJ 1 Calorie 1 cal 10 J = 1.7154 x 10 4 = 1.7 x 10 4 kJ 6.15 Time = () 33 3 4.1 x 10 Cal 10 cal 4.184 J 1 kJ h 1.0lb 1.0 lb 1 Cal 1 cal 1850 kJ 10 J    = 9.2726 = 9.3 hour 6.16 The system does work and thus its internal energy is decreased. This means the sign will be
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/30/2011 for the course CHEM 103 taught by Professor Lewis during the Spring '08 term at Wisc Eau Claire.

### Page1 / 21

Chapter 6 - CHAPTER 6 THERMOCHEMISTRY: ENERGY FLOW AND...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online