595_Physics ProblemsTechnical Physics

595_Physics - Chapter 20 Again K i = 5.00 J Both blocks must rise equally in temperature(c T = Q = mcT Q 5.00 J = = 4.04 10 3 C mc 2 1.60 kg 387 J

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Chapter 20 597 (c) Again, K i = 500 . J. Both blocks must rise equally in temperature. “” Qm c T =∆ : T Q mc == ⋅° ° . . 387 404 10 3 J 21 .60 kg Jkg C C bg b g At any instant, the two blocks are at the same temperature, so for both Q = 0. For the moving block: K =− . J and E int J =+ 250 . so W J For the stationary block: K = 0 and E int J . so W J For each object in each situation, the general continuity equation for energy, in the form ∆∆ KE W Q += + int , correctly describes the relationship between energy transfers and changes in the object’s energy content. P20.67 AA A A A ++ end walls ends of attic side walls roof A A kA T L + × × × ° L N M O Q P + ° F H G I K J = ×⋅ ° ° 2800 22 1 2 400 370 2100 304 480 10 304 0210 17 4 4 15 4 .. t a n . . . m 5.00 m m 4.00 m m 5.00 m m m cos37.0 m kW m C m C m kW kcal s 2 2 a f a f
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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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