In a copper vessel we pour 20 gm of hot water at 80°C . Now 100 gm of cold water at 10°C is added to the vessel. Find the final temperature of mixture if vessel weighs 50 gm.
Answers
Constants that we need:
Heat capacity of water is Cpw = 4.186J/g-K (same for vessel)
The final temperature of the water is denoted as Tf
To raise the temperature of this 80g + 20g = 100g of water and vessel from 20C to Tf requires
ΔH1 = mCpwΔT = 100(4.186)(Tf – 20) = (418.6Tf – 8372)J
To cool 100g of water from 40C to Tf releases
ΔH2 = mCpwΔT = 100(4.186)(40 – Tf) = (16,744 – 418.6Tf)J
But ΔH1 = ΔH2 in a sealed system (no heat flow to or from the outside)
418.6Tf – 8372 = 16744 – 418.6Tf
837.2Tf = 25,116
Tf = 30
The final temperature of the water and vessel is 30C.
P.S. We could just have said that 100g at 20C + 100g at 40C results in 200g at 30C, but the method shown is the correct way to calculate the temperature.
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Answer: 50*0.4(80-T) + 20*4.2(80-T) = 100*4.2(T-10)
=23.89' C maybe
Do the calculation by your own once. If correct mark this as brainliest or comment so that I can also know.
Explanation:
In hot body, for copper m=50g c=0.4 T=80 for water, m=20 C= 4.2 T=80
in cool body, for water m=100g C=4.2 T=10
and W= m C(del T)
By help of Alakh Pandey sir in "Thermal Properties of Matter video 03"