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Three bodies of the same material and having masses $m, m$ and $3 m$ are at temperatures $40^{\circ} \mathrm{C}, \quad 50^{\circ} \mathrm{C} \quad$ and $\quad 60^{\circ} \mathrm{C}$ respectively. If the bodies are brought in thermal contact, the final temperature will be
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$54^{\circ} \mathrm{C}$
Let the final temperature after the masses in thermal contact is $\theta$, then from the principle of calonimetry.
Heat lost $=$ Heat gained
$\begin{aligned} \Rightarrow 3 m s(60-\theta) &=m s(\theta-50)+m s(\theta-40) \\ \Rightarrow \quad 3(60-\theta) &=\theta-50+\theta-40 \\ \Rightarrow \quad 180-3 \theta &=2 \theta-90 \\ \theta &=\frac{270}{5}=54^{\circ} \mathrm{C} \end{aligned}$
Heat lost $=$ Heat gained
$\begin{aligned} \Rightarrow 3 m s(60-\theta) &=m s(\theta-50)+m s(\theta-40) \\ \Rightarrow \quad 3(60-\theta) &=\theta-50+\theta-40 \\ \Rightarrow \quad 180-3 \theta &=2 \theta-90 \\ \theta &=\frac{270}{5}=54^{\circ} \mathrm{C} \end{aligned}$
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