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)`

Options

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))`

Answer

B

Explanation

If the equilibrium temperature `TgtT_(1)andT_(2)` but less than `T_(3)` then as there is no heat loss to the surroundings therfore heat lost loss to the surroundings therfore heat lost by `M_(1)and M_(2)="heat gained by "M_(3)`
`M_(1)s(T_(1)-T_(2))+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`
`rArr" "(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))`