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 temperature $T$ is ( $s$ is specific heat of copper).

T = \frac{T_{1} + T_{2} + T_{3}}{3}

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T = \frac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}

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T = \frac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{3 (M_{1} + M_{2} + M_{3})}

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T = \frac{M_{1} T_{1} s + M_{2} T_{2} s + M_{3} T_{3} s}{M_{1} + M_{2} + M_{3}}

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Solution

The correct option is A $T = \frac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}$
Let us assume that $T_{1} > T_{2}, T_{3}$ and $T_{1} > T > T_{2}, T_{3}$

Now heat loss by $M_{1} =$ Heat gained by $M_{2}$ and $M_{3}$

$M_{1} S (T_{1} - T) = M_{2} S (T - T_{1}) + M_{3} S (T - T_{3})$

$⟹ M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3} = (M_{1} + M_{2} + M_{3}) T$

$⟹ T = \frac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}$

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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 A T=M1T1+M2T2+M3T3M1+M2+M3Let us assume that T1>T2,T3 and T1>T>T2,T3Now heat loss by M1= Heat gained by M2 and M3M1S(T1−T)=M2S(T−T1)+M3S(T−T3)⟹M1T1+M2T2+M3T3=(M1+M2+M3)T⟹T=M1T1+M2T2+M3T3M1+M2+M3