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標題:

Physics conservation of energy

發問:

5kg of steam at 100 deg is mixed with 1 kg of ice at 0 deg.What is the final temperature of the mixture? Assume no heat loss to the surrounding. 更新: Could you please list the steps? thanks

最佳解答:

Latnet heat of fusion of water = 334 000 J/kg Latent heat of vapourization of water = 2260 000 J/kg Heat required to melt 1 kg of ice at 0'C = 334 000 J Heat required to heat 1 kg of water from 0'C to 100'C = 4200 x 100 J = 420 000 J (where 4200 J/kg-K is the specific heat capacity of water) Hence, total heat required to change 1 kg ice at 0'C to 1 kg water at 100'C = (334 000 + 420 000) J = 754 000 J Therefore, mass of steam condensed in order to supply 754 000 J of energy = 754000/2260000 kg = 0.33 kg Since there are initially 5 kg of steam, thus mass of steam left = (5-0.33) kg = 4.67 kg The final mixture consists of 4.67 kg of steam at 100'C, together with (0.33+1) kg = 1.33 kg of water at 100'C Since the steam and water are now at thermal equilibrium (both at 100'C) with each other, no further condensation or evapouration would take place. The final mixture therefore maintains at 100'C The final mixture

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