# PHYS102hw2 - HOMEWORK Ch11 Problems...

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HOMEWORK Ch11 Problems 4,7,11,13,17,22,28,35,37,55 PROBLEM SOLUTIONS 11.4 The change in temperature of the rod is     4 1.00 10 J 31.7°C 0.350 kg 900 J kg°C Q T mc and the change in the length is   0 1 62 24 10 °C 20.0 cm 31.7°C 1.52 10 cm 0.152 mm L L T     11.7 (a)       2 2 2 3 3 net 0 11 75 kg 11.0 m s 0 4.54 10 J 4.5 22 f W KE m   vv (b) 3 net 2 4.54 9.1 10 J s 910 W 5.0 s W t P (c) If the mechanical energy is 25% of the energy gained from converting food energy, then   net 0.25 WQ  and 0.25 ( ) Qt P , so the food energy conversion rate is 910 J s 1 Cal 0.87 Cal s 0.25 0.25 4 186 J Q t    P (d) The excess thermal energy is transported by conduction and convection to the surface of the skin and dis posed of through the evaporation of sweat. 11.11 The quantity of energy transferred from the water-cup combination in a time interval of 1 minute is water cup 3 JJ 0.800 kg 4 186 0.200 kg 900 1.5°C 5.3 kg °C kg °C Q mc mc T  The rate of energy transfer is

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HOMEWORK Ch11 3 5.3 10 J J 88 88 W 60 s s Q t P 11.13 From   0 L L T , the required increase in temperature is found, using Table 10.1, as 3 steel 0 3.0 10 m L T L  1 6 11 10 °C 13 yd 1 yd 3.0 ft 3.281 ft    1 m 23°C The mass of the rail is 70 lb w m g  yd 13 yd   2 4.448 N 9.80 m s 1 lb 2 4.1 10 kg  so the required thermal energy (assuming that steel iron cc ) is         26 steel 4.1 448 J kg °C 23°C 4.2 Q mc T 11.17 The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver sample. Thus,   Al Cu Ag c s w w w m c m c m c T mc T   Solving for the mass of the cup gives Ag Ag Ag Cu Al 1 c s w w w T m m c m c m c cT  , or                 87 32 1 400 g 234 40 g 387 225 g 4 186 80 g 900 32 27 c m 11.22 The kinetic energy given up by the car is absorbed as internal energy by the four brake drums (a total mass of 32 kg of iron). Thus,   drums Fe KE Q m c T or 2 11 2 car 22 drums Fe 1 500 kg 30 m s 47°C 32 kg 448 J kg °C i m T mc v 11.28 The energy required is the following sum of terms:
HOMEWORK Ch11 energy to reach melting point energy to melt energy to reach bo iling point energy to vaporize energy to reach 110°C Q  Mathematically,       ice steam 0°C 10°C 100°C 0°C 110°C 100°C fw Q m c L c L c     v This yields 35 6 JJ 40 10 kg 2 090 10°C 3.33

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## This note was uploaded on 03/07/2011 for the course PHYS 102 taught by Professor Armenn.kocharian during the Winter '11 term at California State University Los Angeles .

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PHYS102hw2 - HOMEWORK Ch11 Problems...

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