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 gt T_2 gt T_3). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)

Three copper blocks of masses M_1, M_2 and M_3 kg respectively are bro

1.	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 gt T_2 gt T_3). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)
Answer» `T=(T_1+T_2+T_3)/3` `T=(M_1T_1+M_2T_2+M_3T_3)/(M_1+M_2+M_3)` `T=(M_1T_1+M_2T_2+M_3T_3)/(3(M_1+M_2+M_3))` `T=(M_1T_1s+M_2T_2s+M_3T_3s)/(M_1+M_2+M_3)` Solution :Suppose temperature of SYSTEM in EQUILIBRIUM is T and `T_1 gt T_2 gt T_3`and suppose, `T_1 LT T, T_2 lt T`and `T_3 gt T`. Block does not lost heat in sink hence heat lost by `M_3`= Heat gained by `M_1` + heat gained by `M_2` `M_3 s(T_3-T)=M_1s(T-T_1)+M_2s(T-T_2)` `therefore M_3sT_3-M_3sT=M_1sT-M_1sT_1 + M_2sT-M_2sT_2` `therefore M_3sT_3+M_1sT_1 + M_2sT_2 = M_1sT+ M_2sT+M_3sT` `therefore M_1T_1+M_2T_2+M_3T_3 =[M_1+M_2+M_3]T` `T=(M_1T_1+M_2T_2+M_3T_3)/(M_1+M_2+M_3)`