Question

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)

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

The value of $\theta $ (in $^{o}C$ to the nearest integer) is___________

Answer 1

Let $\theta _{1}$ , $\theta _{2}$ and $\theta _{3}$ be the temperatures of water in three containers.
$1\left(\theta \right)_{1}+2\left(\theta \right)_{2}+0\left(\theta \right)_{3}=\left(1 + 2\right)60$
$\theta _{1}+2\theta _{2}=180$ …. (1)
$0\times \left(\theta \right)_{1}+1\times \left(\theta \right)_{2}+2\times \left(\theta \right)_{3}=\left(1 + 2\right)30$
$\theta _{2}+2\theta _{3}=90$ …… (2)
$2\times \left(\theta \right)_{1}+0\times \left(\theta \right)_{2}+1\times \left(\theta \right)_{3}=\left(2 + 1\right)60$
$2\theta _{1}+\theta _{3}=180$ …. (3)
and $\left(\theta \right)_{1}+\left(\theta \right)_{2}+\left(\theta \right)_{3}=\left(1 + 1 + 1\right)\theta $
from (1)+(2)+(3)
$3\theta _{1}+3\theta _{2}+3\theta _{3}=450\Longrightarrow \theta _{1}+\theta _{2}+\theta _{3}=150$
from (4) equation $150=3\theta \Longrightarrow \theta =50^\circ C$