552_Physics ProblemsTechnical Physics

552_Physics ProblemsTechnical Physics - 554 P19.15...

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

View Full Document Right Arrow Icon
554 Temperature P19.15 (a) LT L T Al Al Brass Brass 11 +=+ αα ∆∆ bg b g T LL T TT = = ×− × =− ° = − ° −− Al Brass Brass Brass Al Al C so C. This is attainable. 10 01 10 00 10 00 19 0 10 10 01 24 0 10 199 179 66 .. af ej (b) T = × 10 02 10 00 10 00 19 0 10 10 02 24 0 10 T ° 396 C so T =− ° 376 C which is below 0 K so it cannot be reached. P19.16 (a) AA T i = 2 α : A ° ° 2170 10 00800 500 61 2 . C m C A = 109 10 0109 5 m c m 22 (b) The length of each side of the hole has increased. Thus, this represents an increase in the area of the hole. P19.17 VV T i = × − × = βα 3 5 81 10 3 11 0 10 50 0 20 0 0 548 46 . . . gal gal P19.18 (a) T i =+ 1 : 5050 5000 1 240 10 200 . . cm cm C C × ° ° T T 437 C (b) We must get Al Brass = for some T , or L T ii ,, Al Al Brass Brass cm C cm C 1 240 10 1 190 10 += + ° = ° b g Solving for T , T 2080 C, so T 3000 C This will not work because aluminum melts at 660 C ° . P19.19 (a)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 .

Ask a homework question - tutors are online