633_Physics ProblemsTechnical Physics

633_Physics ProblemsTechnical Physics - Chapter 22 *P22.5...

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Chapter 22 635 *P22.5 (a) The input energy each hour is 789 10 2500 60 118 10 39 .. ×= × Jrevo lu t ion revm in min 1 h Jh ej bg implying fuel input 1 29 4 9 × × F H G I K J = L 4.03 10 J Lh 7 (b) QW Q hc =+ eng . For a continuous-transfer process we may divide by time to have Q t W t Q t W t Q t Q t h c h c ∆∆∆ ∆∆ == = × × F H G I K J F H G I K J = eng eng eng Useful power output J revolution J revolution 1 min min 60 s W W 1 hp 746 W hp 7 89 10 4 58 10 2500 rev 1 138 10 185 33 5 5 . . P (c) eng eng Js rev 60 s rev 2 r ad Nm =⇒ = = × F H G I K J =⋅ τω τ ωπ 1 527 5 . (d) Q t c = × F H G I K J 458 10 191 10 3 5 . . J revolution rev 60 s W P22.6 The heat to melt 15.0 g of Hg is Qm L cf × × = 15 10 1 18 10 177 34 kg J kg J e j . The energy absorbed to freeze 1.00 g of aluminum is L hf × = 10 3 97 10 397 35 kg J / kg
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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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