579_Physics ProblemsTechnical Physics

579_Physics ProblemsTechnical Physics - Chapter 20 P20.10 b...

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Chapter 20 581 P20.10 (a) fm g h m cT bg b g =∆ 0600 300 10 980 500 4186 300 00924 0760 258 3 .. . . . .; . af ej e j b g a f × =⋅ ° = ° kg m s m Jca l g c a l g C CC 2 T TT (b) No . Both the change in potential energy and the heat absorbed are proportional to the mass; hence, the mass cancels in the energy relation. *P20.11 We do not know whether the aluminum will rise or drop in temperature. The energy the water can absorb in rising to 26 ° C is mc T ∆= ° °= 025 6 6279 . k g J kg C C J . The energy the copper can put out in dropping to 26 ° mc T ° 0 1 387 74 2 864 . kg J kg C C J . Since 6 279 2 864 J J > , the final temperature is less than 26 ° C. We can write QQ hc =− as Q T T T T ff f f f f water Al Cu kg J kg C C k g J kg C C kg J kg C C C C C ++ = ° −°+ ° −° + ° = +− ° + ° = 0 0 25 20 0 4 900 26 01 387 100 0 1 046 5 20 930 360 9 360 38 7 3 870 0 1 445 2 34160 23 6 . . . di P20.12 cold hot m cTT m cTT mc m c T mc m c T mcT mcT m c mc mc T mc T mcT
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