Q.Three copper blocks of masses $M_1$, $M_2$ and $M_3$ kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at $T_1$, $T_2$, $T_3$ ($T_1 > T_2 > T_3$). Assuming there is no heat loss to the surroundings, the equilibrium temprature $T$ is ($s$ is specific heat of copper)

Concept: Specific Heat Capacity — when bodies of the same material are brought into thermal contact, heat flows until all reach a common equilibrium temperature. Since no heat is lost, the total heat gained equals the total heat lost.

Step 1: For each block, the heat change is $Q = M s (T_f - T_i)$. Because $s$ is the same for all (copper), it cancels out.

Step 2: Heat lost by hotter blocks = heat gained by cooler blocks. Writing the energy balance:

Step 3: Cancel $s$ and solve for $T$: …

When three objects at different temperatures are brought into thermal contact, heat flows from the hotter ones to the colder ones until they all reach a common temperature $T$. The key principle is conservation of energy: the total heat lost by the blocks that cool down equals the total heat gained by the blocks that warm up, provided no heat escapes to the surroundings.

Here, all three blocks are made of the same material (copper), so they share the same specific heat capacity $s$. That means the heat required to change the temperature of a block by $\Delta T$ is $Q = M s \Delta T$. Because $s$ is identical for all, it will cancel out — a fact that immediately eliminates options that keep $s$ in the numerator (like option D) or that ignore mass altogether (like option A).

Let’s work through the calculation step by step.

1. Write the heat lost and gained.

   Block 1 is the hottest ($T_1 > T$), so it loses heat:

   $Q_{\text{lost}} = M_1 s (T_1 - T)$.

   Block 3 is the coldest ($T > T_3$), so it gains heat:

   $Q_{\text{gained,3}} = M_3 s (T - T_3)$.

   Block 2 is intermediate — it could either lose or gain heat depending on where $T$ falls relative to $T_2$. But we don’t need to decide; the algebra will handle it automatically if we write every block’s heat change as $M s (T_{\text{final}} - T_{\text{initial}})$ and sum to zero.

2. Apply conservation of energy.

   The net heat exchanged among the three blocks is zero:

Notice that each term is $M s$ times the temperature change. If a block cools, $(T - T_{\text{initial}})$ is negative, correctly representing heat lost.

3. Factor out the common $s$. Since $s \neq 0$, we can divide the entire equation by $s$: $M_1 (T - T_1) + M_2 (T - T_2) + M_3 (T - T_3) = 0.$ …

Recognize this as a centre-of-mass problem in disguise: since the heat capacity of each block is $M_is$, the equilibrium temperature is the mass-weighted mean of the initial temperatures, $T=\dfrac{\sum M_iT_i}{\sum M_i}$, exactly like locating a centre of mass with 'masses' $M_i$ sitting at 'positions' $T_i$. Two quick checks confirm option (b) without deriving it: if $M_1=M_2=M_3$ it must reduce to the plain average $(T_1+T_2+T_3)/3$, which only (b) does; and $T$ must lie strictly between $T …