MCQ
Three containers $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in litres) are taken from each containers and mixed (assume no loss of heat during the process)

$\begin{array}{|c|c|c|c|}\hline \mathrm{C_{1 }} & {\mathrm{C}_{2}} & {\mathrm{C}_{3}} & {\mathrm{T}} \\ \hline {1 l} & {2 l} & {-} & {60^{\circ} \mathrm{C}} \\ \hline {-} & {1 l} & {2 l} & {30^{\circ} \mathrm{C}} \\ \hline  {2 l} & {-} & {1 l} & {60^{\circ} \mathrm{C}} \\ \hline  {1 l} & {1 l} & {1 l} & {\theta} \\ \hline\end{array}$

The value of $\theta$ (in $^{\circ} \mathrm{C}$ to the nearest integer) is

  • A
    $45$
  • B
    $48$
  • C
    $55$
  • $50$

Answer

Correct option: D.
$50$
d
According to table and applying law of calorimetry

$1 \mathrm{T}_{1}+2 \mathrm{T}_{2}=(1+2) 60^{\circ}$

$=180$

$1 \mathrm{T}_{2}+2 \mathrm{T}_{3}=(1+2) 30^{\circ}$

$=90$

$2 \mathrm{T}_{1}+1 \mathrm{T}_{3}=(1+2) 60$

$=180$

$3\left(\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}\right)=450$

$\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}=150^{\circ}$

Hence,

$\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}=(1+1+1) \theta$

$150=3 \theta$

$\theta=50^{\circ} \mathrm{C}$

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