標題:
f3 physic 222~~~PLX ~~~
發問:
A cooker supplies steam at 110℃ at the rate of 0.1kg per min/. caculate the rate,in watts,at which the cooker supplies heat energy if the steam condenses to water at 100℃....specific heat capacity of steam is 2000Jkg-1℃-1
最佳解答:
First of all, calculate the energy lost by the steam at 110℃ to 100℃. Mass of steam in one minute, m = 0.1 kg Specific heat capacity of steam, c = 2000 Jkg^-1℃^-1 Temperature change of steam, ΔT = (110 - 100) = 10℃ Assume no energy is lost to the surroundings, By the law of conservation of energy, Energy lost by steam = mcΔT = (0.1)(2000)(10) = 2000 J Then, we have to calculate the amount of energy lost when steam condenses to water. Specific latent heat of vaporization of water, lv = 2.26 X 10^6 Jkg^-1 Assume no energy is lost to the surroundings, By the law of conservation of energy, Energy lost = mlv = (0.1)(2.26 X 10^6) = 2.26 X 10^5 J ∴ Total energy lost in 1 minute = 2000 + (2.26 X 10^5) = 2.28 X 10^5 J ∴ Rate of supply of energy by the cooker = Energy / time = (2.28 X 10^5) / 60 = 3800 Js^-1 = 3800 W
其他解答:
2000*0.1*(110-100)+2.26*10^6*0.1=P*60 2000*0.1*(110-100) is the energy require to 110oC steam change into 100oC steam 2.26*10^6*0.1 is the energy require to 100oC steam change into 100oC water P*60 is the total energy require. the answer is 3800W.|||||P = E/t = 0.1 (2000) (110-100) / 60 = 33.33W
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