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 wf + F H G I K J = bullet where m w is the melt water mass m m w w = ×+ × ° ° × = + = −− 0500300 10 240 300 10 300 333 10 86 4 11 5 0294 3 2 3 5 .. . . . . kg m s kg 128 J kg C C Jkg J J 333 000 J kg g ej bg b g af P20.18 (a) Q 1 = heat to melt all the ice × = × 50 0 10 3 33 10 1 67 10 35 4 . kg J kg J e j Q 2 34 50 0 10 4186 100 2 09 10 ° ° = × heat to raise temp of ice to 100 C kg J kg C C J a f Thus, the total heat to melt ice and raise temp to 100°C 376 10 4 . J Q 3 36 4 10 0 10 2 26 10 2 26 10 == × × = × heat available as steam condenses kg J kg J . e j Thus, we see that QQ 31 > , but QQQ 312 <+ . Therefore, all the ice melts but T f 100 C. Let us now find T f 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 0 10 0 10 2 26 10 10 0 10 100 3 3 . . e j di e j From which, T f 40 4 .C . (b) Q 1 = heat to melt all ice 167 10 4 J
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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 .

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