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200 gm of water at 20 degree Celsius is mixed with 20 gm of steam at120 degree Celsius find mixture temperature.the process is not carried out in atmospheric pressure hence the boiling point of water is 110degree celius

200 gm of water at 20 degree Celsius is mixed with 20 gm of steam at120 degree Celsius find mixture temperature.the process is not carried out in atmospheric pressure hence the boiling point of water is 110degree celius

Grade:11

1 Answers

Kruthik
119 Points
3 years ago
Let T be the final temperature of the mixture.
Let’s calculate the amount of heat released by steam to be converted into water at ToC.
First, steam at 120will be converted into 110o
Heat relesed during this process Q1=mS∂T=20g×0.46cal g-1 oC-1 × 10oC = 92cal.
Heat released during the conversion of steam in water Q2=mLv=20g × 540calg-1 = 10780cal
Heat released during the conversion of water at 110oC to water at ToC = mS∂T = 20g×1calg-1oC-1(110 – T)
Now Qsteam=Q+Q+ Q= (11092 + 20T)
Now let’s calculate heat absorbed by water to get convert to ToC
Qwater = mS∂T = 200g × 1cal g-1 oC-1× (T-20) = 200T – 4000.
By principle of Calorimetry, Qsteam = Qwater.
Hence, 11092 + 20T = 200T – 4000
So, T = 63.84oC.
 
            

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