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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_1T_1+M_2T_2+M_3T_3)/(M_1+M_2+M_3)`

C

`T=(M_1T_1+M_2T_2+M_3T_3)/(3(M_1+M_2+M_3))`

D

`T=(M_1T_1s+M_2T_2s+M_3T_3s)/(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 lt T lt T_3`.
According to question, there is no heat loss to the surroundings.
Heat lost by `M_3` = Heat gained by `M_1` + Heat gained by `M_2`
`implies 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)
`implies T[M_1 +M_2 +M_3] =M_3T_3 + M_1T_1 +M_2T_2`
`implies T= (M_1T_1 + M_2T_2 +M_3T_3)/(M_1 +M_2 +M_3)`
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