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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 reach equlibrium. 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))/(3(M_(1)+M_(2)+M_(3))`

C

`T= (M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(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
If the equilibrium tepm. `T gt T_(1)` and `T_(2)` but less than `T_(3)`, then as there is no heat loss to the surroundings, therfore, heat lost by `M_(1)` and `M_(2)` = heat gained by `M_(3)`
`M_(1)s(T_(1)-T)+M_(2)s(T_(2)-T)=M_(3)s(T-T_(3))`
`M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3)= (M_(1)+M_(2)+M_(3))T`
or `T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))`
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