hw14_sol_s11

hw14_sol_s11 - Phys 2101 Homework 14 Solution Spring `11...

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Phys 2101 Homework 14 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1.The change of diameter is . 00627 . 0 ) 000 . 0 0 . 100 )( / 10 23 )( 725 . 2 ( 6 cm C C C T D D T D D o o o Al 2. If V c is the original volume of the cup, a is the coefficient of linear expansion of aluminum, and T is the temperature increase, then the change in the volume of the cup is V c = 3 a V c T . See Eq. 18-11. If is the coefficient of volume expansion for glycerin, then the change in the volume of glycerin is V g = V c T . Note that the original volume of glycerin is the same as the original volume of the cup. The volume of glycerin that spills is           4 6 3 3 3 5.1 10 / C 3 23 10 / C 100cm 6.0 C 0.26cm . g c a c V V V T           Note: Glycerin spills over because 3 ba > , which gives 0 gc VV  . Note that since liquids in general have greater coefficients of thermal expansion than solids, heating a cup filled with liquid generally will cause the liquid to spill out. 3. The mass m = 0.100 kg of water, with specific heat c = 4190 J/kg·K, is raised from an initial temperature T i = 23°C to its boiling point T f = 100°C. The heat input is given by Q = cm ( T f T i ). This must be the power output of the heater P multiplied by the time t ; Q = Pt . Thus,     () 4190J/kg K 0.100kg 100 C 23 C 160s. 200J/s fi cm T T Q t PP 4. The melting point of silver is 1235 K, so the temperature of the silver must first be raised from 15.0° C (= 288 K) to 1235 K. This requires heat 4 ( ) (236J/kg K)(0.130kg)(1235 C 288 C) 2.91 10 J. Q cm T T   Now the silver at its melting point must be melted. If L F is the heat of fusion for silver, this requires     34 0.130kg 105 10 J/kg 1.36 10 J. F Q mL The total heat required is ( 2.91 10 4 J + 1.36 10 4 J ) = 4.27 10 4 J. 5. Let the mass of the steam be m s and that of the ice be m i . Then ( 0.0 C) (100 C ) F c w c f s s s w f L m c m T m L m c T   , where T f = 50°C is the final temperature. We solve for m s : ( 0.0 C) (79.7cal / g)(150g) (1cal / g· C)(150g)(50 C 0.0°C) (100 C ) 539cal / g (1cal / g C )(100 C 50 C) 33g.
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This note was uploaded on 05/09/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.

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hw14_sol_s11 - Phys 2101 Homework 14 Solution Spring `11...

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