{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

581_Physics ProblemsTechnical Physics

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}