Thermodynamics HW Solutions 70

Thermodynamics HW Solutions 70 - 42 1 ( ) ( ) [ ] ( ) [ ] C...

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Chapter 1 Basics of Heat Transfer 1-132 A 0.2-L glass of water at 20°C is to be cooled with ice to 5°C. The amounts of ice or cold water that needs to be added to the water are to be determined. Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water is negligible. Properties The density of water is 1 kg/L, and the specific heat of water at room temperature is C = 4.18 kJ/kg·°C (Table A-9). The heat of fusion of ice at atmospheric pressure is 333.7 kJ/kg,. Analysis The mass of the water is () () kg 0.2 L 0.2 kg/L 1 = = ρ = V m w We take the ice and the water as our system, and disregard any heat and mass transfer. This is a reasonable assumption since the time period of the process is very short. Then the energy balance can be written as ( ) ( ) 0 0 water ice energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net = Δ + Δ Δ = Δ = U U U E E E out in 43 42 1 43
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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)(5-0) C] + (0.2 kg)(4.18 kJ/kg C)(5-20) 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)(5-20) 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. 1-70...
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