Three copper blocks of masses M1, M2 and M3 kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T1, T2, T3 (T1 > T2 > T3 ). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)

The correct option is:
T = (M1T1+M2T2+M3T3)/(M1+M2+M3)

Step-by-Step Explanation:

1. Understanding the Principle of Calorimetry:
According to the principle of calorimetry, in an isolated system with no heat exchange with the surroundings, the total heat lost by the hotter bodies must be equal to the total heat gained by the colder bodies. Mathematically, the sum of all heat exchanges within the system is zero:
$Q_1+Q_2+Q_3=0$
where $Q_i$ is the heat transferred to or from the i-th block.

2. Expressing Heat Exchange for Each Block:
The heat transferred to or from a block of mass M, specific heat capacity s, initial temperature $T_i$, and final equilibrium temperature T is given by:
$Q=M·s·(T-T_i)$
Since all three blocks are made of copper, they share the same specific heat capacity s. We can write the heat equations for the three blocks as:
$Q_1=M_1·s·(T-T_1)$
$Q_2=M_2·s·(T-T_2)$
$Q_3=M_3·s·(T-T_3)$

3. Applying the Conservation of Energy:
Substitute the expressions for $Q_1$, $Q_2$, and $Q_3$ into the conservation equation:
$M_{1}s(T-T_1)+M_{2}s(T-T_2)+M_{3}s(T-T_3)=0$

4. Solving for Equilibrium Temperature T:
Since the specific heat capacity s is a non-zero common factor, we can divide the entire equation by s:
$M_1(T-T_1)+M_2(T-T_2)+M_3(T-T_3)=0$
Expanding the terms:
$M_{1}T-M_1{T}_1+M_{2}T-M_2{T}_2+M_{3}T-M_3{T}_3=0$
Group the terms containing T on one side:
$T(M_1+M_2+M_3)=M_1{T}_1+M_2{T}_2+M_3{T}_3$
Divide both sides by $(M_1+M_2+M_3)$ to find the equilibrium temperature:
$T=\frac{M_1{T}_1+M_2{T}_2+M_3{T}_3}{M_1+M_2+M_3}$