Fourth HW Solution - C HAPTER 10 Problem Solutions 10.1 (a)...

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Unformatted text preview: 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 = + =- + =- + = 2 8 0 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 =-- = 6 2 .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 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 .2 9 g a l of gasoline will overflow....
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This note was uploaded on 06/19/2008 for the course PHYS 221 taught by Professor Ezell during the Summer '08 term at Campbell University .

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Fourth HW Solution - C HAPTER 10 Problem Solutions 10.1 (a)...

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