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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
  • JEE Mains 2020
  • Hard
  • Numerical

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

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Answer & Solution

Correct Answer

(D) \(50\)

Step-by-step Solution

Detailed explanation

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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