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Three copper blocks of masses $M_1, M_2$ and $M_3 \mathrm{~kg}$ respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at $T_1, T_2, T_3\left(T_1>\right.$ $\mathrm{T}_2>\mathrm{T}_3$ ). Assuming there is no heat loss to the surroundings, the equilibrium temperature $T$ is ( $s$ is specific heat of copper)
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Verified Answer
The correct answer is:
$T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3}$
$T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3}$
Consider the equilibrium temperature of the system is $T$.
Let us consider, $T_1, T_2 < T < T_3$.
As given that, there is no net loss to the surroundings. Heat lost by $M_3=$ Heat gained by $M_1$
+ Heat gained by $M_2$
$$
\begin{aligned}
&M_3 s\left(T_3-T\right)=M_1 s\left(T-T_1\right)+M_2 s\left(T-T_2\right) \\
&M_3 s T_3-M_3 s T=M_1 s T-M_1 s T_1+M_2 s T-M_2 s T_2
\end{aligned}
$$
(where, $s$ is specified heat of the copper material)
$$
\begin{aligned}
&\quad T\left[M_1+M_2+M_3\right]=M_3 T_3+M_1 T_1+M_2 T_2 \\
&T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3}
\end{aligned}
$$
Let us consider, $T_1, T_2 < T < T_3$.
As given that, there is no net loss to the surroundings. Heat lost by $M_3=$ Heat gained by $M_1$
+ Heat gained by $M_2$
$$
\begin{aligned}
&M_3 s\left(T_3-T\right)=M_1 s\left(T-T_1\right)+M_2 s\left(T-T_2\right) \\
&M_3 s T_3-M_3 s T=M_1 s T-M_1 s T_1+M_2 s T-M_2 s T_2
\end{aligned}
$$
(where, $s$ is specified heat of the copper material)
$$
\begin{aligned}
&\quad T\left[M_1+M_2+M_3\right]=M_3 T_3+M_1 T_1+M_2 T_2 \\
&T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3}
\end{aligned}
$$
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