JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 \(55\)
- D \(50\)
Answer & Solution
Correct Answer
(D) \(50\)
Step-by-step Solution
Detailed explanation
According to table and applying law of calorimetry \(1 \mathrm{T}_{1}+2 \mathrm{T}_{2}=(1+2) 60^{\circ}\) \(=180\) \(1 \mathrm{T}_{2}+2 \mathrm{T}_{3}=(1+2) 30^{\circ}\) \(=90\) \(2 \mathrm{T}_{1}+1 \mathrm{T}_{3}=(1+2) 60\) \(=180\)…
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