Three liquids with masses m_1, m_2, m_3 are t | vedclass

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(B) According to the principle of calorimetry,the total heat lost by the hotter liquids is equal to the total heat gained by the colder liquids.Let $T

Vedclass · Accepted Answer

$\frac{m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3}{m_1 c_1 + m_2 c_2 + m_3 c_3}$

Vedclass · Answer

(B) According to the principle of calorimetry,the total heat lost by the hotter liquids is equal to the total heat gained by the colder liquids.Let $T$ be the final equilibrium temperature of the mixture.The heat gained or lost by a substance is given by $Q = mc\Delta T$.For the mixture,the total heat exchange must be zero:$m_1 c_1 (T - T_1) + m_2 c_2 (T - T_2) + m_3 c_3 (T - T_3) = 0$Expanding the terms:$m_1 c_1 T - m_1 c_1 T_1 + m_2 c_2 T - m_2 c_2 T_2 + m_3 c_3 T - m_3 c_3 T_3 = 0$Grouping the terms with $T$:$T(m_1 c_1 + m_2 c_2 + m_3 c_3) = m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3$Solving for $T$:$T = \frac{m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3}{m_1 c_1 + m_2 c_2 + m_3 c_3}$

Three liquids with masses $m_1, m_2, m_3$ are thoroughly mixed. If their specific heats are $c_1, c_2, c_3$ and their temperatures are $T_1, T_2, T_3$ respectively,then the temperature of the mixture is

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