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f.3 physics

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A 1-kW heater,immersed in water(0.5,20度)is switched on for 10min.Calculate the maximum amount of water boiled away.

最佳解答:

Assume no heat is lost to the surroundings, Energy supply by the heater = Power×Time = 1000×(10×60) = 600 000 J Energy required to raise the temperature of water from 20℃ to 100℃ = mc△T = (0.5)(4200)(100–20) = 168000 J ∴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 = 2.26×106 Jkg-1 432 000 = m(2.26×106) m = 0.191 kg ∴the mass of water boiled away is 0.191 kg.

其他解答:

maximum and Adobe Reader 8????481517FD598DAD6B

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