Question

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)

पर्याय

• `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)`

Answer 1

`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)`