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In a container of negligible mass, ‘m’ g...

In a container of negligible mass, ‘m’ grams of steam at 100°C is added to 100 g of water that has temperature 20°C. If no heat is lost to the surroundings at equilibrium, match the items given in Column I with that in Column II.

Text Solution

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The correct Answer is:
A-P;B-P;C-SD-Q
If Q is the heat required to convert 100g of water at 20°C to 100°C
Then `Q=msDeltatheta=100(1)(100-20)orQ=8000" cal"`
If `m_(0)` = mass of steam that converts into water to liberate this much amount of heat,
`m_(0)=(Q)/(L_(V))=(8000)/(540)=14.8gms`
Since, it is less than m = 20g, the temperature of mixture is 100°C
Mass of steam in mixture = (20 – 14.8) = 5.2 gms
Mass of water in mixture = (100 + 14.8) = 114.8 gms
If m = 10g, amount of heat liberated by steam `=10xx540=5400`cal
Let `theta` be the final temperature of mixture, `m_(H_(2)OS_(H_(2))O(theta-20)=m_("steam")L_(V)+mS_(H_(2))O(100-theta)rArr100xx1(theta-20)=10xx540+10(1)(100-theta)`
or `110theta=5400+1000+2000rArrtheta=76.4^(@)C`
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