{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol to assignment10

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

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

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}