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10分：A F. 3 Physics Question (Heat)

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發問:

10分：A F. 3 Physics Question (Heat) A 1-kW heater, immersed in water (0.5 kg, Temperature=20 degree Celsius), is switched on for 10 minutes. Calculate the maximum amount of water boiled away. Please show your working steps clearly and explain your answer briefly.

最佳解答:

Assume no heat is lost to the surroundings, Energy supply by the heater = Power X Time = 1000 X (10 X 60) = 600 000 J Energy required to raise the temperature of water from 20*C to 100*C = mc△T = (0.5)(4200)(100 - 20) = 168 000 J So, energy used for boiling the water = 600 000 – 168 000 = 432 000 J E = mL, where L is the specific latent heat of vaporization of water = 2.26 X 106 Jkg-1 432 000 = m(2.26 X 106) m = 0.191 kg So, the mass of water boiled away is 0.191 kg.

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