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Three copper blocks of masses M(1), M(2)...

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))`

Text Solution

Verified by Experts

The correct Answer is:
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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