ssm_ch18 - Chapter 18 Student Solutions Manual 8. The...

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Chapter 18 – Student Solutions Manual 8. The change in length for the aluminum pole is 6 01 (33m)(23 10 / C )(15 C)= 0.011m. A T α Δ= Δ = × ° ° AA 15. If V c is the original volume of the cup, a is the coefficient of linear expansion of aluminum, and Δ T is the temperature increase, then the change in the volume of the cup is Δ V c = 3 a V c Δ T . See Eq. 18-11. If β is the coefficient of volume expansion for glycerin then the change in the volume of glycerin is Δ V g = V c Δ 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 () ( ) ( ) ( ) 46 3 3 3 5.1 10 / C 3 23 10 / C 100cm 6.0 C 0.26cm . gc a c VV V T βα −− ⎡⎤ Δ− Δ= − × ° × ° ° ⎣⎦ = 21. Consider half the bar. Its original length is 00 /2 L = A and its length after the temperature increase is T =+ Δ A . The old position of the half-bar, its new position, and the distance x that one end is displaced form a right triangle, with a hypotenuse of length A , one side of length , and the other side of length x . The Pythagorean theorem yields Since the change in length is small we may approximate (1 + Δ T ) 0 A 2222 22 0 (1 ) . x =−= + Δ A A T 2 by 1 + 2 Δ T , where the small term ( Δ T ) 2 was neglected. Then, 2 2 2 0 0 x TT Δ −= Δ αα and 62 0 3.77m 2 2 25 10 /C 32 C 7.5 10 m. 2 xT = × ° ° = × A 25. 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. fi Qc m 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 Qm L == × = × The total heat required is ( 2.91 × 10 4 J + 1.36 × 10 4 J ) = 4.27 × 10 4 J.
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27. 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 ⋅° ° == = = 41. (a) We work in Celsius temperature, which poses no difficulty for the J/kg·K values of specific heat capacity (see Table 18-3) since a change of Kelvin temperature is numerically equal to the corresponding change on the Celsius scale. There are three possibilities: • None of the ice melts and the water-ice system reaches thermal equilibrium at a temperature that is at or below the melting point of ice.
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This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.

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ssm_ch18 - Chapter 18 Student Solutions Manual 8. The...

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