HW12_Sol_F10

HW12_Sol_F10 - Nov 15 2010 PHYS 2101 HW#12 WileyPlus...

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PHYS 2101 HW#12 WileyPlus Problem Solutions 1 HW #12 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by []) 1) [18-7] We assume scale X is a linear scale in the sense that if its reading is x then it is related to a reading y on the Kelvin scale by a linear relationship y = mx + b . We determine the constants m and b by solving the simultaneous equations: 373.15 = m ( ! 53.5) + b & 273.15 = m ( ! 170) + b which yield the solutions m = 100/(170 – 53.5) = 0.858 and b = 419. With these values, we find x for y = 340: 2) [18-13] Since a volume is the product of three lengths, the change in volume due to a temperature change ! T is given by ! V = 3 ! V ! T , where V is the original volume and is the coefficient of linear expansion. See Eq. 18-11. Since V = (4 " /3) R 3 , where R is the original radius of the sphere, then The value for the coefficient of linear expansion is found in Table 18-2. The change in volume can be expressed as , where is the coefficient of volume expansion. For aluminum, we have . 3) [18-17] 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 ! V g " ! V c = # " 3 \$ a ( ) V c ! T = 5.1 % 10 " 4 / C ° ( ) " 3 23 % 10 " 6 / C ° ( ) ' ( ) 100cm 3 ( ) 6.0 ° C ( ) = 0.26cm 3 . Note: Glycerin spills over because , which gives . 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. 4) [18-23] 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, 5) [18-27] 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 Now the silver at its melting point must be melted. If L F is the heat of fusion for silver, this requires The total heat required is ( 2.91 # 10 4 J + 1.36 # 10 4 J ) = 4.27 # m i . Then ,

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HW12_Sol_F10 - Nov 15 2010 PHYS 2101 HW#12 WileyPlus...

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