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

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Let θ _ 1 , θ _ 2 and θ _ 3 be the temperatures of water in three containers. 1 θ _ 1 + 2 θ _ 2 + 0 θ _ 3 = (1 + 2)60 θ _ 1 + 2 θ _ 2 = 180 &hellip;. (1) 0 × θ _ 1 + 1 × θ _ 2 + 2 × θ _ 3 = (1 + 2)30 θ _ 2 + 2 θ _ 3 = 90 &hellip;&hellip; (2) 2 × θ _ 1 + 0 × θ _ 2 + 1 × θ _ 3 = (2 + 1)60 2 θ _ 1 + θ _ 3 = 180 &hellip;. (3) and θ _ 1 + θ _ 2 + θ _ 3 = (1 + 1 + 1)θ from (1)+(2)+(3) 3 θ _ 1 + 3 θ _ 2 + 3 θ _ 3 = 450 ⟹ θ _ 1 + θ _ 2 + θ _ 3 = 150 from (4) equation 150 = 3θ ⟹ θ = 50 ° C

${C}_{1}$	${C}_{2}$	${C}_{3}$	$T$
$1l$	$2l$	$- -$	${60}^{o}C$
$- -$	$1l$	$2l$	${30}^{o}C$
$2l$	$- -$	$1l$	${60}^{o}C$
$1l$	$1l$	$1l$	$\theta$

Three containers C _ 1 , C _ 2 and C _ 3 have water at different...

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