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प्रश्न
Three copper blocks of masses M1, M2 and M3 kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T1, T2, T3 (T1 > T2 > T3). Assuming there is no heat loss to the surroundings, the equilibrium temprature T is (s is specific heat of copper)
विकल्प
`T = (T_1 + T_2 + T_3)/3`
`T = (M_1T_1 + M_2T_2 + M_3T_3)/(M_1 + M_2 + M_3)`
`T = (M_1T_1 + M_2T_2 + M_3T_3)/(3(M_1 + M_2 + M_3))`
`T = (M_1T_1s + M_2T_2s + M_3T_3s)/(M_1 + M_2 + M_3)`
उत्तर
`T = (M_1T_1 + M_2T_2 + M_3T_3)/(M_1 + M_2 + M_3)`
Explanation:
Let the equilibrium temperature to be found be T. Now, we consider T to be greater than T1 and T2 but smaller than T3.
Since there is no loss of heat energy. Hence, we get,
Heat lost by M3 = Heat regained by M1 + Heat regained by M2 So, we get,
⇒ M3s (T3 – T) = M1s (T – T1) + M2s (T – T2)
Dividing both sides by s, we get,
⇒ M3 (T3 – T) = M1 (T – T1) + M2 (T – T2)
Opening the brackets and solving for T, we get,
⇒ M3T3 – M3T = M1T – M1T1 + M2T – M2T2
⇒ M3T3 + M1T1 + M2T2 = M1T + M2T + M3T
⇒ `T = (M_1T_1 + M_2T_2 + M_3T_3)/(M_1 + M_2 + M_3)`
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