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Question

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 the specific heat of copper).

A
T=T1+T2+T33
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B
T=m1T1+m2T2+m3T3m1+m2+m3
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C
T=m1T1S+m2T2S+m3T3Sm1+m2+m3
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D
T=m1T1+m2T2+m3T33(m1+m2+m3)
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Solution

The correct option is B T=m1T1+m2T2+m3T3m1+m2+m3
Heat loss by m1 = Heat gained by m2 + Heat gained by m3

So, m1S(T1T)=m2S(TT2)+m3S(TT3)

m1T1m1T=m2Tm2T2+m3Tm3T3

T(m1+m2+m3)=m1T1+m2T2+m3T3

T=m1T1+m2T2+m3T3m1+m2+m3

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