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  • Three copper blocks of masses M_1, M_2 and M_ 3 kg respectively are brought into thermal contact till they reach equilibrium.

Three copper blocks of massesM_{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} > T_{2} > T_{3} ). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)

a) T=\frac{T_{1}+T_{2}+ T_{3}}{3}

b) T=\frac{M_{1}T_{1}+M_{2}T_{2}+ M_{3}T_{3}}{M_{1}+M_{2}+M_{3}}

c) T=\frac{M_{1}T_{1}+M_{2}T_{2}+ M_{3}T_{3}}{3(M_{1}+M_{2}+M_{3})}

d) T=\frac{M_{1}T_{1}s+M_{2}T_{2}s+ M_{3}T_{3}s}{M_{1}+M_{2}+M_{3}}

Answers (1)

The answer is the option (b) Let the equilibrium temperature of the system=T

Let T_{1}T_{2}<T<T_3

As there is no loss to the surroundings.

Heat lost by M_3=Heat gain by M_{1}+Heat gain by M_{2}

M_{3}s(T_{3}-T)=M_{1}s(T-T_{1})+M_{2}s(T-T_{2})

M_3sT_3-M_3sT=M_1sT-M_1sT_1+M_2sT-M_2sT_2

  T(M_{3}+M_{1}+M_2)=M_3T_3+M_1T_1+M_2T_2

 T=\frac{M_1T_1+M_2T_2+M_{3}T_{3}}{M_{1}+M_2+M_3}

Hence (b) is correct.

Posted by

infoexpert24

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