Three containers C_{1}, C_{2} and C_{3} have | vedclass

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(D) Let the temperatures of water in containers $C_{1}, C_{2}$,and $C_{3}$ be $T_{1}, T_{2}$,and $T_{3}$ respectively.Using the principle of calorimet

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(D) Let the temperatures of water in containers $C_{1}, C_{2}$,and $C_{3}$ be $T_{1}, T_{2}$,and $T_{3}$ respectively.Using the principle of calorimetry $(m_{1}T_{1} + m_{2}T_{2} = (m_{1}+m_{2})T_{mix})$:$1$. For the first mixture: $1T_{1} + 2T_{2} = (1+2)60 = 180$  ---$(i)$$2$. For the second mixture: $1T_{2} + 2T_{3} = (1+2)30 = 90$  ---(ii)$3$. For the third mixture: $2T_{1} + 1T_{3} = (2+1)60 = 180$  ---(iii)Adding equations $(i)$,(ii),and (iii):$(1+2)T_{1} + (2+1)T_{2} + (2+1)T_{3} = 180 + 90 + 180$$3(T_{1} + T_{2} + T_{3}) = 450$$T_{1} + T_{2} + T_{3} = 150^{\circ} C$For the final mixture of $1 \ l$ each:$1T_{1} + 1T_{2} + 1T_{3} = (1+1+1)	heta$$150 = 3	heta$$	heta = 50^{\circ} C$

$C_{1}$	$C_{2}$	$C_{3}$	$T$
$1 \ l$	$2 \ l$	$-$	$60^{\circ} C$
$-$	$1 \ l$	$2 \ l$	$30^{\circ} C$
$2 \ l$	$-$	$1 \ l$	$60^{\circ} C$
$1 \ l$	$1 \ l$	$1 \ l$	$\theta$

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