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Heat

Three copper ...

Question

Three copper blocks of masses $M_{1}, M_{2}, and M_{3} k g$ 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)

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

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

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

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T = \frac{T_{1} + T_{2} + T_{3}}{3}

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Solution

The correct option is B $T = \frac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}$
As per the given question, since there is no heat loss to the surroundings and the equilibrium temperature of the system is $T$ .
As we know, specific heat is given by,
$Q = M s Δ T$
Where $s$ is specific heat of material.
Therefore,
Heat lost by $M_{3}$ = $Heat gained by M_{1} + Heat gained by M_{2}$

Heat lost by $M_{3} = Heat gained by M_{1} + Heat gained by M_{2}$

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

(Here, $s$ is the specific heat of the copper material)

$T [M_{1} + M_{2} + M_{3}] = M_{3} T_{3} + M_{1} T_{1} + M_{2} T_{2}$

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

Correct Answer: (b).

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Heat

Standard XII Physics

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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 temperature T is (𝑠 is specific heat of copper)