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If 20 g of steam at 100 oC is mixed into 80 g of water at...

If 20 g of steam at 100 oC is mixed into 80 g of water at 20 oC, what will be the final temperature if no thermal energy is lost and no steam escapes?

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Answer #1

Solution: mass of steam = m1 = 20g.

mass of water = m2 = 80g.

Temperature of steam = T1 = 1000c.

Temperature of water = T2 = 200c.

Specific heat of stam = 4.187J/g-K.

specific heat of steam = 1.996J/g-K.

latent heat of steam = 2257J/g.

Let equilibrium temperature = T.

Energy released when steam converted into water = E1 = 20*2257 = 45140J.

Energy required during temperature increase of water from 200c to 1000c = E2 = 80*4.187*80 = 9516.8J.

Since E1 is greater than E2, hence water temperature become 1000c.

  

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