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標題:
phy ~heat(15 points)
how to do it ? The question:Milk power is mixed with 0.17(kg) hot water at 61(。c) in a milk bottle. Mary finds the milk is too hot so sheputs the bottle in a water bath of 20(。c) to cool down the milk. The temperature of milk drops to 35(。c) after 5 mins.Assume there is no energy loss to the... 顯示更多 how to do it ? The question: Milk power is mixed with 0.17(kg) hot water at 61(。c) in a milk bottle. Mary finds the milk is too hot so sheputs the bottle in a water bath of 20(。c) to cool down the milk. The temperature of milk drops to 35(。c) after 5 mins. Assume there is no energy loss to the surroundings. (specific heat capacity of milk/ water = 4200) 1 .How much energy does the milk lose to the water in the bath? 2 . What is the mass of water(water bath)? 3. What is the rate of energy transfer from the milk to the water? Please give me the steps , not only the ans. thx^_^
最佳解答:
1. E = mc delta T = 0.17 x 4200 x (61 - 35) = 18564 J 2. energy lost by the milk = energy gained by the water bath m1x c1 x delta T1 = m2 x c2 x delta T2 0.17 x 4200 x ( 61 - 35) = m2 x 4200 x (35 - 20) m2 = 0.295 kg 3. rate of energy transfer from the milk to the water: P = E/T P = 18564/ (5 x 60s) = 61.88 W
其他解答:
phy ~heat(15 points)
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發問:how to do it ? The question:Milk power is mixed with 0.17(kg) hot water at 61(。c) in a milk bottle. Mary finds the milk is too hot so sheputs the bottle in a water bath of 20(。c) to cool down the milk. The temperature of milk drops to 35(。c) after 5 mins.Assume there is no energy loss to the... 顯示更多 how to do it ? The question: Milk power is mixed with 0.17(kg) hot water at 61(。c) in a milk bottle. Mary finds the milk is too hot so sheputs the bottle in a water bath of 20(。c) to cool down the milk. The temperature of milk drops to 35(。c) after 5 mins. Assume there is no energy loss to the surroundings. (specific heat capacity of milk/ water = 4200) 1 .How much energy does the milk lose to the water in the bath? 2 . What is the mass of water(water bath)? 3. What is the rate of energy transfer from the milk to the water? Please give me the steps , not only the ans. thx^_^
最佳解答:
1. E = mc delta T = 0.17 x 4200 x (61 - 35) = 18564 J 2. energy lost by the milk = energy gained by the water bath m1x c1 x delta T1 = m2 x c2 x delta T2 0.17 x 4200 x ( 61 - 35) = m2 x 4200 x (35 - 20) m2 = 0.295 kg 3. rate of energy transfer from the milk to the water: P = E/T P = 18564/ (5 x 60s) = 61.88 W
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