CalorimetryMedium
Question
5g of steam at 100oC is mixed at 100oC is mixed with 10g of ice at 0oC. Chose correct alternative/s) :- (Given swater = 1 cal / goC, LF = 80 cal / g, LV = 540 cal / g)
Options
A.Equilibrium ,temperature of mixture is is 160oC
B.Equilibrium ,temperature of mixture is is 100oC
C.At equilibrium, mixture contain 13
g of water
g of water D.At equilibrium, mixture contain 1
g of steam
g of steamSolution
Required heat Available heat
10 g ice (0oC) 5 g steam (100oC)
↓800 cal ↓ 2700 cal
10 g water (0oC) 5 g water (100oC)
↓1000 cal
10 g water (100o)
So available heat is more than required heat therefore final temperature will be 100oC.
Mass of vapour condensed
=
g
Total mass of water
= 10 +
= 13
g
Total mass of steam
= 5 -
g
10 g ice (0oC) 5 g steam (100oC)
↓800 cal ↓ 2700 cal
10 g water (0oC) 5 g water (100oC)
↓1000 cal
10 g water (100o)
So available heat is more than required heat therefore final temperature will be 100oC.
Mass of vapour condensed
=
gTotal mass of water
= 10 +
= 13 gTotal mass of steam
= 5 -
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