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Three copper blocks of masses M(1), M(2)...

Three copper blocks of masses `M_(1), M_(2) and M_(3)` kg respectively are brought into thermal contact till they each equilibrium. 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)

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))`

Text Solution

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The correct Answer is:
To find the equilibrium temperature \( T \) of the three copper blocks, we can follow these steps: ### Step 1: Understand the Heat Transfer When the three copper blocks are brought into thermal contact, heat will flow from the hotter blocks to the cooler blocks until they reach thermal equilibrium. Since there is no heat loss to the surroundings, the heat lost by the hotter blocks will equal the heat gained by the cooler blocks. ### Step 2: Set Up the Heat Transfer Equation Let: - \( M_1 \) be the mass of the first block, initially at temperature \( T_1 \) - \( M_2 \) be the mass of the second block, initially at temperature \( T_2 \) - \( M_3 \) be the mass of the third block, initially at temperature \( T_3 \) Given that \( T_1 > T_2 > T_3 \), we can express the heat lost and gained as follows: - Heat lost by block 1: \( Q_1 = M_1 s (T_1 - T) \) - Heat lost by block 2: \( Q_2 = M_2 s (T_2 - T) \) - Heat gained by block 3: \( Q_3 = M_3 s (T - T_3) \) ### Step 3: Apply the Conservation of Energy According to the principle of conservation of energy, the total heat lost by blocks 1 and 2 should equal the total heat gained by block 3: \[ Q_1 + Q_2 = Q_3 \] Substituting the expressions for heat: \[ M_1 s (T_1 - T) + M_2 s (T_2 - T) = M_3 s (T - T_3) \] ### Step 4: Simplify the Equation Since \( s \) (specific heat) is a common factor, we can cancel it out from the equation: \[ M_1 (T_1 - T) + M_2 (T_2 - T) = M_3 (T - T_3) \] Expanding this gives: \[ M_1 T_1 - M_1 T + M_2 T_2 - M_2 T = M_3 T - M_3 T_3 \] ### Step 5: Rearrange the Equation Rearranging the terms to isolate \( T \): \[ M_1 T_1 + M_2 T_2 + M_3 T_3 = M_1 T + M_2 T + M_3 T \] Combining the terms involving \( T \): \[ M_1 T_1 + M_2 T_2 + M_3 T_3 = (M_1 + M_2 + M_3) T \] ### Step 6: Solve for the Equilibrium Temperature \( T \) Now, we can solve for \( T \): \[ T = \frac{M_1 T_1 + M_2 T_2 + M_3 T_3}{M_1 + M_2 + M_3} \] This gives us the equilibrium temperature of the three blocks. ### Final Answer The equilibrium temperature \( T \) is: \[ T = \frac{M_1 T_1 + M_2 T_2 + M_3 T_3}{M_1 + M_2 + M_3} \]
To find the equilibrium temperature \( T \) of the three copper blocks, we can follow these steps: ### Step 1: Understand the Heat Transfer When the three copper blocks are brought into thermal contact, heat will flow from the hotter blocks to the cooler blocks until they reach thermal equilibrium. Since there is no heat loss to the surroundings, the heat lost by the hotter blocks will equal the heat gained by the cooler blocks. ### Step 2: Set Up the Heat Transfer Equation Let: - \( M_1 \) be the mass of the first block, initially at temperature \( T_1 \) ...
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