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sol to assignment10

sol to assignment10 - .4 Swat/W 5” ““77"” 11.7...

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Unformatted text preview: .4 Swat/W 5” ““77"”; 11.7 Considera 1. 0011ng of water: (1= APE: mgh= (1. 00 kg)(9. 80 m/s2 Also, A T= 3‘ 490 mc 30. R=10.1C°(100 ““4186 x 103 J/k2°C) =0117° C )(50. 0111): =4901 11.13 (heat gain = heat loss) becomes: ' mwaterch 11,5191ng AT. Thus, mam-(4186 J/kg“C)(25.'0°C)= (3.00 kg)(129 J/kg°C)( 500°C), or Water’- 0.185 kg= 185 g. ' 11.21 heat gaiu= heat loss . (heattomeltice) +(Heattowarmme1tediceto Tf) =(loss ofheat by 1 kg or water) ..(0 100 kg)(3.33 x105 J/kg) + (0.100 kg)(4186 J/kg°C)( Tf) '. = (1.00 kg)(4186 J/kg"C)(80.0 ~ T5) = 655°C (66°C) '_9 1.1211 (11.8 -l—)(o 16 m2)(11 1°12) - A -s m°C I 11.3‘1(a)_HeM.=-1—=‘_ 30 x 10-3 m . =47ol. (into house) ‘ (0.3 —.L)(o 16 m2)(33. 9°C) (b) H—Ag—M— C 1.7 103 1- ‘At'L ‘ Q 3.0x10-3m' '= x “‘3 outdoors) - ‘ " ' ' 11 37 The area of the 81mm p ‘ . - =4atr2 1):; 312:6]:- 76:). and fora perfecjggigfig? ):=‘1‘52x10'2 . x10 W/m2K4N4-52X10'2m2)((,473—K4Thus =110W.. ) '(295 K)4) \ 11.49 The power incident 'onthe solar collector is . Pi = IA = (600 W/m2)u(0.25 m)2 = 117.8 w. - . - ‘ - . For a 50% reflector, the collected. power is Pc— — 53. 9 W. The total energy required to increase the temperature of the water to the boiling point and to evaporate it is Q: Inch 1+ va= (1.0 kg)[(4186 J/lcg’C)(80°C) + (2.26 x 106 J/kg)] or of. 2. 59 x 106 J. 6 The time required is At=§=1538§ 119s =‘4.41x1043= 12 h. ...
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