Three containers C 1 , C 2 and C 3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)
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Sol. 1 θ 1 + 2 θ 2 = (1 + 2) 60
θ 1 + 2 θ 2 = 180 ….
0 × θ 1 + 1 × θ 2 + 2 × θ 3 = (1 + 2) 30
θ 2 + 2 θ 3 = 90 ….
2 × θ 1 + 0 × θ 2 + 1 × θ 3 = (2 + 1) 60
2 θ 1 + θ 3 = 180 ….
and θ 1 + θ 2 + θ 3 = (1 + 1 + 1) θ ….
from + + ls
3 θ 1 + 3 θ 2 + 3 θ 3 = 450 ⇒ θ 1 + θ 2 + θ 3 = 150
from equation 150 = 3 θ ⇒ θ = 50ºC
ls 150 = 3 θ ⇒ θ = 50ºC
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