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Fourth HW Solution - CHAPTER 10 Problem Solutions 10.1(a TF...

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C HAPTER 10 Problem Solutions 10.1 (a) ( 29 F C 9 9 32 273.15 32 5 5 T T = + = - + = -460 F ° (b) ( 29 ( 29 C F 5 5 32 98.6 32 9 9 T T = - = - = 37.0 C ° (c) ( 29 ( 29 F C K 9 9 9 32 273.15 32 173.15 32 5 5 5 T T T = + = - + = - + = 280 F - ° 10.4 ( 29 ( 29 C F 5 5 32 134 32 9 9 T T = - = - = 56.7 C ° and ( 29 C 5 79.8 32 9 T = - - = 62.1 C - ° 10.14 The desired change in volume is ( 29 3 3 4 3 6 3 1 m 100 cm 1.00 10  m 10  cm V - = = × From  ( 29 ( 29 ( 29 3 i i V V T V T β α = = , the required change in temperature is ( 29 ( 29 ( 29 4 3 1 6 3 1.00 10  m 3 3 24 10   C 1.00 m i V T V α - - - × = = = × ° 1.39 C ° 1
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C H A P T E R   1 0 10.21 The initial volume of the gasoline is  45 L i V = . As the temperature rises to 35 ° C, this volume  will expand by ( 29 ( 29 ( 29 ( 29 1 4 9.6 10   C 45 L 35 C 10 C 1.1 liters i V V T β - - = = × ° ° - ° = . Thus, if the tank does not expand,  ( 29 1.1 L   0.29 gal  of gasoline will overflow. C HAPTER 11 Problem Solutions 11.1  The Gravitational Potential Energy is the source of the heat energy ( 29 Q mc T = = mgh     c gh T ÷ = C C kg J m s m 0 0 2 1 . 0 ) / 4186 /( ) 0 . 50 )( / 8 . 9 ( = T T T + = 1
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