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Unformatted text preview: 42 1 ( ) ( ) [ ] ( ) [ ] C C water 1 2 ice liquid 2 solid 1 = + + + T T mC T mC mh T mC if o o Water 0.2 L 20 C Ice, C Noting that T 1, ice = 0 C and T 2 = 5 C and substituting m [0 + 333.7 kJ/kg + (4.18 kJ/kg C)(50) C] + (0.2 kg)(4.18 kJ/kg C)(520) C = 0 It gives m = 0.0354 kg = 35.4 g Cooling with cold water can be handled the same way. All we need to do is replace the terms for ice by the ones for cold water at 0 C: ( ) ( ) ( ) [ ] ( ) [ ] water 1 2 water cold 1 2 water water cold = + = + T T mC T T mC U U Substituting, [ m cold water (4.18 kJ/kg C)(5  0) C] + (0.2 kg)(4.18 kJ/kg C)(520) C = 0 It gives m = 0.6 kg = 600 g Discussion Note that this is 17 times the amount of ice needed, and it explains why we use ice instead of water to cool drinks. 170...
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 Fall '10
 Dr.DanielArenas
 Thermodynamics, Mass, Heat

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