# Ch19Word - Chapter 19 CHAPTER 19 Heat and the First Law of...

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Chapter 19 Page 1 CHAPTER 19 – Heat and the First Law of Thermodynamics 1. The required heat flow is ? Q = mc ? T = (30.0 kg)(4186 J/kg · C°)(95°C – 15°C) = 1.0 × 10 7 J . 2. We find the temperature from ? Q = mc ? T ; 7700 J = (3.0 kg)(4186 J/kg · C°)( T – 10.0°C), which gives T = 10.6°C . 3. The heat flow generated must equal the kinetic energy loss: ? Q = – ( ! mv f 2 ! mv i 2 ) = ! m ( v i 2 v f 2 ) = ! (3.0 × 10 –3 kg)[(400 m/s) 2 – (200 m/s) 2 ] = 1.8 × 10 2 J . 4. We convert the units: ? Q = mc ? T ; 1 Btu = (1 lb)(1.00 kcal/kg · C°)(1 F°)(0.454 kg/lb)(5 C°/9 F°) = 0.252 kcal; 1 Btu = (0.252 kcal)(4186 J/kcal) = 1055 J. 5. We find the mass per hour from ? Q / t = ( m / t ) c ? T ; 7200 kcal/h = ( m / t )(1.00 kcal/kg · C°)(50°C – 15°C), which gives m / t = 2.1 × 10 2 kg/h . 6. We find the time from ? Q = mc ? T ; (350 W) t = (250 mL)(1.00 g/mL)(10 –3 kg/g)(4186 J/kg · C°)(60°C – 20°C), which gives t = 120 s . 7. The heat flow generated must equal the kinetic energy loss: ? Q = ! mv 2 = ! (1000 kg)[(95 km/h)/(3.6 ks/h)] 2 (1 kcal/4186 J) = 83 kcal . 8. We find the specific heat from ? Q = mc ? T ; 135 × 10 3 J = (5.1 kg) c (30°C – 20°C), which gives c = 2.6 × 10 3 J/kg · . 9. The required heat flow is ? Q = mc ? T = (16 L)(1.00 kg/L)(4186 J/kg · C°)(90°C – 20°C) = 4.7 × 10 6 J . 10. We find the temperature from heat lost = heat gained; m water c water ? T water = m glass c glass ? T glass ; (135 mL)(1.00 g/mL)(10 –3 kg/g)(4186 J/kg · C°)( T – 39.2°C) = (0.035 kg)(840 J/kg · C°)(39.2°C – 21.6°C), which gives T = 40.1°C . 11. If all the kinetic energy in the hammer blows is absorbed by the nail, we have K = 10( ! mv 2 ) = mc ? T ; 10[ ! (1.20 kg)(6.5 m/s) 2 ] = (0.014 kg)(450 J/kg · C°) ? T , which gives ? T = 40 C° . 12. We find the temperature from heat lost = heat gained; m Cu c Cu ? T Cu = ( m Al c Al + m water c water ) ? T Al ; (0.245 kg)(390 J/kg · C°)(300°C – T ) = [(0.150 kg)(900 J/kg · C°) + (0.820 kg)(4186 J/kg · C°)]( T – 12.0°C), which gives T = 19.5°C .

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Chapter 19 Page 2 13. We find the temperature from heat lost = heat gained; m shoe c shoe ? T shoe = ( m pot c pot + m water c water ) ? T pot ; (0.40 kg)(450 J/kg · C°)( T – 25°C) = [(0.30 kg)(450 J/kg · C°) + (1.35 L)(1.00 kg/L)(4186 J/kg · C°)](25°C – 20°C), which gives T = 186°C . 14. We find the specific heat from heat lost = heat gained; m x c x ? T x = ( m Al c Al + m water c water + m glass c glass ) ? T water ; (0.215 kg) c x (330°C – 35.0°C) = [(0.100 kg)(900 J/kg · C°) + (0.150 kg)(4186 J/kg · C°) + (0.017 kg)(840 J/kg · C°)](35.0°C – 12.5°C), which gives c x = 260 J/kg · . 15. The water must be heated to the boiling temperature, 100°C. We find the time from t = (heat gained)/ P = [( m Al c Al + m water c water ) ? T water ]/ P = [(0.360 kg)(900 J/kg · C°) + (0.75 L)(1.00 kg/L)(4186 J/kg · C°)](100°C – 8.0C)/(750 W) = 425 s = 7.1 min . 16. The Calorie content of 100 g of brownie will be 10 times the thermal energy released when 10 g is ignited: Q = 10(heat gained) = 10( m bomb c Al + m water c water + m cup c Al ) ? T = 10[(0.615 kg)(0.22 kcal/kg · C°) + (2.00 kg)(1.00 kcal/kg · C°) + (0.524 kg)(0.22 kcal/kg · C°)](36.0°C – 15.0°C) = 470 kcal . 17. ( a ) The heat required at a temperature T to raise the temperature by a small amount d T is d Q = mc ( T ) d T . We add the total heat required to raise the temperature from T 1 to T 2 by integrating: Q =d Q 0 Q = mc ( T )d T T 1 T 2 .
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Ch19Word - Chapter 19 CHAPTER 19 Heat and the First Law of...

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