InterviewSolution
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InterviewSolution
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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 each equilibrium. Before contact, they were at `T_(1),T_(2),T_(3) (T_(1)gtT_(2)gtT_(3))`. Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)A. `T=(T_(1)+T_(2)+T_(3))/(3)`B. `T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))`C. `T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(3(M_(1)+M_(2)+M_(3)))`D. `T=(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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Answer» Correct Answer - B Let the equilibrium temperature of the system is T. Let us assume that `T_(1),T_(2) ltTltT_(3)`. ltBrgt According to question, there is no net loss to the surroundings. Heat lost by `M_(3)`= Heat gained by ` M_(1)` + Heat gained by `M_(2)` ltBrgt `rArr" "M_(3)s(T_(3)-T)=M_(1)s(T-T_(1))+M_(2)s(T-T_(2))` `" "` (where, s is specific heat of the copper material) `rArr" "T[M_(1)+M_(2)+M_(3)]=M_(3)T_(3)+M_(1)T_(1)+M_(2)T_(2)` `rArr" "T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))` |
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