Chapter 20579SOLUTIONS TO PROBLEMSSection 20.1Heat and Internal EnergyP20.1Taking m=100.kg, we have∆Umghg===1 009 8050 0490...kgm smJ2bgejaf.But∆∆∆UQmcTTg=⋅°=4186490. kgJkgCJbgso ∆T=°0117.CTT Tfi=+ =+°∆10 0 0 117afCP20.2The container is thermally insulated, so no energy flows by heat:Q=0and∆EQWWmghint=+=inputinput02The work on the falling weights is equal to the work done on thewater in the container by the rotating blades. This work results inan increase in internal energy of the water:2mghEmc Tintwater∆Tmghmc×⋅°=°22 1 509 803 000 20088 20105water2kgm smkgJ kg CJ837 J CC....FIG. P20.2Section 20.2Specific Heat and CalorimetryP20.3QmcT=silver12305250234..kJkgCkJ kg Csilversilver=⋅°ccP20.4FromcT=∆
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .