화학공정계산 3판 솔루션 RICHARD M.FELDER Elementary principles of chemical processes 솔루션

화학공정계산 3판 솔루션 RICHARD M.FELDER Elementary principles of chemical processes 솔루션

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화학공정계산 3판 솔루션 RICHARD M.FELDER Elementary principles of chemical processes

화학공정계산 3판 솔루션 RICHARD M.FELDER - Elementary principles of chemical processes

화학공정계산 3판 솔루션 RICHARD M.FELDER - Elementary principles of chemical processesCHAPTER TWO
2.1 (a)
= 18144 × 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h ⇒ 26.0 mi / h 3.2808 ft 1 h 3 wk 7d 24 h 3600 s 1000 ms

(c)

554 m 4 1d 1h d ? kg 24 h 60 min

1 kg 108 cm 4 = 3.85 × 10 4 cm 4 / min? g 1000 g 1 m 4

2.2 (a) (b) (c)

1 m 1 h 760 mi = 340 m / s h 0.0006214 mi 3600 s 921 kg 2.20462 lb m m3 1 kg 1 m3 = 57.5 lb m / ft 3 35.3145 ft 3 1.34 × 10 -3 hp = 119.93 hp ⇒ 120 hp 1 J/s

5.37 × 10 3 kJ 1 min 1000 J min 60 s 1 kJ

2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 in cube. For a classroom with dimensions 40 ft × 40 ft × 15 ft : 40 × 40 × 15 ft 3 (12) 3 in 3 1 ball n balls = . = 518 × 10 6 ? 5 million balls ft 3 2 3 in 3 The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4.3 light yr 365 d 24 h
1 yr 1d 3600 s 1.86 × 10 5 mi 1 h 1 s 3.2808 ft 0.0006214 mi 1 step = 7 × 1016 steps 2 ft

2.5 Distance from the earth to the moo

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