2 3 25 10 mL 1 g water 4 glasses of g water 1 glass 1 mL water The

2 3 25 10 ml 1 g water 4 glasses of g water 1 glass 1

This preview shows page 30 - 34 out of 51 pages.

2 3 2.5 10 mL 1 g water 4 glasses of water 1.0 10 g water 1 glass 1 mL water The heat needed to raise the temperature of 1.0 10 3 g of water is: q ms t (1.0 10 3 g)(4.184 J/g  C)(37 3) C 1.4 10 5 J 1.4 10 2 kJ 3 311.3 kJ/mol H   o 218 2 137.0 kJ/mol H   o 1 174.3 kJ/mol H   o C 2 H 4 + H 2 C 2 H 6 C 2 H 2 + 2H 2
Image of page 30
CHAPTER 6: THERMOCHEMISTRY (b) We need to calculate both the heat needed to melt the snow and also the heat needed to heat liquid water form 0 C to 37 C (normal body temperature). The heat needed to melt the snow is: 2 2 1 mol 6.01 kJ (8.0 10 g) 2.7 10 kJ 18.02 g 1 mol The heat needed to raise the temperature of the water from 0 C to 37 C is: q ms t (8.0 10 2 g)(4.184 J/g  C)(37 0) C 1.2 10 5 J 1.2 10 2 kJ The total heat lost by your body is: (2.7 10 2 kJ) (1.2 10 2 kJ) 3.9 10 2 kJ 6.117 We assume that when the car is stopped, its kinetic energy is completely converted into heat (friction of the brakes and friction between the tires and the road). Thus, 2 1 2 q mu Thus the amount of heat generated must be proportional to the braking distance, d : d q d u 2 Therefore, as u increases to 2 u , d increases to (2 u ) 2 4 u 2 which is proportional to 4 d . 6.118 (a) f f 2 f f (F ) (H O) [ (HF) (OH )]          H H H H H o o o o H [(1)( 329.1 kJ/mol) (1)( 285.8 kJ/mol)] [(1)( 320.1 kJ/mol) (1)( 229.6 kJ/mol) H 65.2 kJ/mol (b) We can add the equation given in part (a) to that given in part (b) to end up with the equation we are interested in. HF( aq ) OH ( aq )   F ( aq ) H 2 O( l ) H 65.2 kJ/mol H 2 O( l )   H ( aq ) OH ( aq ) H 56.2 kJ/mol HF( aq )   H ( aq ) F ( aq ) H 9.0 kJ/mol 6.119 Water has a larger specific heat than air. Thus cold, damp air can extract more heat from the body than cold, dry air. By the same token, hot, humid air can deliver more heat to the body. 6.120 The equation we are interested in is the formation of CO from its elements. C(graphite) 1 2 O 2 ( g )   CO( g ) H ? Try to add the given equations together to end up with the equation above. C(graphite) O 2 ( g )   CO 2 ( g ) H 393.5 kJ/mol CO 2 ( g )   CO( g ) 1 2 O 2 ( g ) H 283.0 kJ/mol 219
Image of page 31
CHAPTER 6: THERMOCHEMISTRY C(graphite) 1 2 O 2 ( g )   CO( g ) H 110.5 kJ/mol 220
Image of page 32
CHAPTER 6: THERMOCHEMISTRY We cannot obtain D H f for CO directly, because burning graphite in oxygen will form both CO and CO 2 . 6.121 Energy intake for mechanical work: 5 3000 J 0.17 500 g 2.6 10 J 1 g 2.6 10 5 J mgh 1 J 1 kg m 2 s 2 2 5 2 2 kg m 2.6 10 (46 kg)(9.8 m/s ) s h h 5.8 10 2 m 6.122 (a) mass 0.0010 kg Potential energy mgh (0.0010 kg)(9.8 m/s 2 )(51 m) Potential energy 0.50 J (b) 2 1 Kinetic energy 0.50 J 2 mu 2 1 (0.0010 kg) 0.50 J 2 u u 2 1.0 10 3 m 2 /s 2 u 32 m/s (c) q ms t 0.50 J (1.0 g)(4.184 J/g C) t t 0.12 C 6.123 For Al: (0.900 J/g
Image of page 33
Image of page 34

You've reached the end of your free preview.

Want to read all 51 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture