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Three copper blocks of masses M1, M2 and...

Three copper blocks of masses `M_1, M_2` and `M_3` kg respectively are brought into thermal contact till they reach equilibrium. Before contact they were at `T_1, T_2, T_3 (T_1 gt T_2 gt T_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
Suppose temperature of system in equilibrium is T and `T_1 gt T_2 gt T_3` and suppose, `T_1 lt T, T_2 lt T` and `T_3 gt T`. Block does not lost heat in sink hence heat lost by `M_3`= Heat gained by `M_1` + heat gained by `M_2`
`M_3 s(T_3-T)=M_1s(T-T_1)+M_2s(T-T_2)`
`therefore M_3sT_3-M_3sT=M_1sT-M_1sT_1 + M_2sT-M_2sT_2`
`therefore M_3sT_3+M_1sT_1 + M_2sT_2 = M_1sT+ M_2sT+M_3sT`
`therefore M_1T_1+M_2T_2+M_3T_3 =[M_1+M_2+M_3]T`
`T=(M_1T_1+M_2T_2+M_3T_3)/(M_1+M_2+M_3)`
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