20gm water of 10°C is filled in a vessel of water equivalent 10 gm, 60gm of water 40°C is added in the beaker then find out final temperature of the mixture.
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Let’s say the final temperature of mixture is x and (20<x<40)
Now Simply Balance the Energy,
Energy Lost by the hot water=Energy Gain by the cold water
mass of hot water*c*(Temp. of hot water-x)=mass of cold water*c*(x-Temp. of cold water)
50*c*(40-x)=75*c*(x-20)
50*(40-x)=75*(x-20)
2*40–2*x=3*x-3*20
5*x=80+60
x=140/5
x=28
so, the final temperature of mixture is 28 degree Celsius
c: Specific Heat capacity of water
Now Simply Balance the Energy,
Energy Lost by the hot water=Energy Gain by the cold water
mass of hot water*c*(Temp. of hot water-x)=mass of cold water*c*(x-Temp. of cold water)
50*c*(40-x)=75*c*(x-20)
50*(40-x)=75*(x-20)
2*40–2*x=3*x-3*20
5*x=80+60
x=140/5
x=28
so, the final temperature of mixture is 28 degree Celsius
c: Specific Heat capacity of water
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