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581_Physics ProblemsTechnical Physics

581_Physics ProblemsTechnical Physics - Chapter 20 P20.17...

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Chapter 20 583 P20.17 The bullet will not melt all the ice, so its final temperature is 0°C. Then 1 2 2 mv mc T m L w f + F H G I K J = bullet where m w is the melt water mass m m w w = × + × ⋅° ° × = + = 0 500 3 00 10 240 3 00 10 30 0 3 33 10 86 4 11 5 0 294 3 2 3 5 . . . . . . . . kg m s kg 128 J kg C C J kg J J 333 000 J kg g e j b g b ga f P20.18 (a) Q 1 = heat to melt all the ice = × × = × 50 0 10 3 33 10 1 67 10 3 5 4 . . . kg J kg J e je j Q 2 3 4 50 0 10 4 186 100 2 09 10 = ° = × ⋅° ° = × heat to raise temp of ice to 100 C kg J kg C C J b g e j b ga f . . Thus, the total heat to melt ice and raise temp to 100°C = × 3 76 10 4 . J Q 3 3 6 4 10 0 10 2 26 10 2 26 10 = = × × = × heat available as steam condenses kg J kg J . . . e je j Thus, we see that Q Q 3 1 > , but Q Q Q 3 1 2 < + . Therefore, all the ice melts but T f < ° 100 C. Let us now find T f Q Q T T f f cold hot kg J kg kg J kg C C kg J kg kg J kg C C = − × × + × ⋅° ° = − × × × ⋅° ° 50 0 10 3 33 10 50 0 10 4 186 0 10 0 10 2 26 10 10 0 10 4 186 100 3 5 3 3 6 3 . . . . . . e je j e j b gd i e je j e j b gd i From which, T f = ° 40 4 . C . (b) Q 1 = heat to melt all ice
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