Physics regarding heat

用戶79880

2007-12-14 5:31 am

1kg steam at 100C is mixed with 1 kg ice at 0C, assumed there is no heat loses to surroundings, what will be the final temperature. In what state will be the result outcome? (Take specific heat capacity as 4200 kg J, take latent of vapourization as 2.26*10^6 and fusion as 3.34*10^5)

回答 (1)

STEVIE-G™

2007-12-14 8:00 am

✔ 最佳答案

The energy released by 100C steam changes to 100C water
=m(lv)
=1(2.26*10^6)
=2.26*10^6 J
The energy required which 0C ice changes to 100C water
=m(lf)+mc(deltaT)
=3.34*10^5+(4200)(100)
=754000 J < 2.26*10^6 J
Therefore,all the ice are changed to 100C water
The mass of 100C steam changes to 100C water
=754000/2.26*10^6
=0.334kg
That means that there remains (1-0.334)=0.666kg steam and the temperature of this 0.666kg steam will not be changed as all of the mixture at this moment are 100C .
Hence,the mixture consists of 100C water and 100C steam .As a result ,the state is the mixture of steam and water

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