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APSC%20170%20Tutorial%2020080227

# APSC%20170%20Tutorial%2020080227 - APSC 175 Matter and...

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Unformatted text preview: APSC 175 Matter and Energy Instructor: Dr. Vladan Prodanovic January — April 2007 School of Engineering, University of British Columbia Okanagan SOLUTIONS: 017.9: It is correct! 17.18: d + Ad = d(1+ 021T) , = (0.4500cm)(1 + (2.4 x 10"5(C°)'1)(23.0°C — (—78.0°C))) = 0.4511 cm = 4.511 mm. 17.28: (a ) No, the brass expands more than the steel. (b) call Do the inside diameter of the steel cylinder at 20°C At 150°C: DsT = D1312 D, + ADST = 25.000 cm + ADE,R Do +aSTD.AT = 25 cm+ aBR(25 cm)AT 25 cm(1+ aBRAT) D0 = —— 1+aﬂAT _ (25 cm)|1 + (2.0 x10‘5(C°)")(130C°)| 1+ (1.2 x10'5(C°)'])(130 C“) = 25.026 cm 17.37: (210° c — 20°c)((1.60 kg)(910 J/kg- K)+ (0.30 kg)(470 J/ kg -K))= 3.03 x 105 1. 21-: McAT _ (70.0 kg)(3480 J/kg- K)(1.00 K) _ 101 L, " , (2.42x106 J/kg) ‘ gay b) This much water has a volume of 101 ems, about a third of a can of soda. 17.54: a) mswem = 17.58: The heat lost by the sample is the heat gained by the calorimeter and water, and the heat capacity of the sample is _ Q _ ((0.200 kg)(4190 J/kg-K)+ (0.150 kg)(390 J/kg —K))(7.1C°) ' mAT " (0.0850 kg)(73.9C°) C =1010 J/kg- K, or 1000 J/ kg- K to the two ﬁgures to which the temperature change is known. 17.70: a) The heat current will be the same in both metals; since the length of the copper rod is known, H =(385.0 W/m-K)(400x10"‘ mzﬁg= 5.39 w. b) The length of the steel rod may be found by using the above value of H in Eq. (17.21) and solving for L2 , or, since H and A are the same for the rods, (50.2 w/m-K)(65.0 K) (385.0 W/m- K1350 K) 12 = LEA—Tl: (1.00m) = 0.242 . k AT m APSC 175 Assignment-01 VP.doc 2 Prodanovic 01/07 ...
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