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 = − + − = Δ + Δ 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...
View
Full Document
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Energy, Mass, Heat, 0.2kg, 0.2L

Click to edit the document details