Three copper blocks of masses $m_{1}$ , $m_{2}$ , 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 the specific heat of copper).

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

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Solution

The correct option is B $T = \frac{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}{m_{1} + m_{2} + m_{3}}$
Heat loss by $m_{1}$ = Heat gained by $m_{2}$ + Heat gained by $m_{3}$

So, $m_{1} S (T_{1} - T) = m_{2} S (T - T_{2}) + m_{3} S (T - T_{3})$

$m_{1} T_{1} - m_{1} T = m_{2} T - m_{2} T_{2} + m_{3} T - m_{3} T_{3}$

$T (m_{1} + m_{2} + m_{3}) = m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}$

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

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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 the specific heat of copper).

The correct option is B T=m1T1+m2T2+m3T3m1+m2+m3Heat loss by m1 = Heat gained by m2 + Heat gained by m3 So, m1S(T1−T)=m2S(T−T2)+m3S(T−T3) m1T1−m1T=m2T−m2T2+m3T−m3T3 T(m1+m2+m3)=m1T1+m2T2+m3T3 ∴ T=m1T1+m2T2+m3T3m1+m2+m3