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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» <html><body><p>`T=(T_1+T_2+T_3)/<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>`<br/>`T=(M_1T_1+M_2T_2+M_3T_3)/(M_1+M_2+M_3)`<br/>`T=(M_1T_1+M_2T_2+M_3T_3)/(3(M_1+M_2+M_3))` <br/>`T=(M_1T_1s+M_2T_2s+M_3T_3s)/(M_1+M_2+M_3)` </p>Solution :Suppose temperature of <a href="https://interviewquestions.tuteehub.com/tag/system-1237255" style="font-weight:bold;" target="_blank" title="Click to know more about SYSTEM">SYSTEM</a> in <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> is T and `T_1 gt T_2 gt T_3`and suppose, `T_1 <a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a> 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` <br/> `M_3 s(T_3-T)=M_1s(T-T_1)+M_2s(T-T_2)` <br/> `therefore M_3sT_3-M_3sT=M_1sT-M_1sT_1 + M_2sT-M_2sT_2` <br/> `therefore M_3sT_3+M_1sT_1 + M_2sT_2 = M_1sT+ M_2sT+M_3sT` <br/> `therefore M_1T_1+M_2T_2+M_3T_3 =[M_1+M_2+M_3]T` <br/> `T=(M_1T_1+M_2T_2+M_3T_3)/(M_1+M_2+M_3)`</body></html>


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