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

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(B) Let the final equilibrium temperature of the mixture be $T$.According to the principle of calorimetry,the total heat lost by the hotter liquids is

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) Let the final equilibrium temperature of the mixture be $T$.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.Alternatively,the sum of heat changes for all components must be zero: $\sum m_i c_i (T - T_i) = 0$.Expanding this,we get: $m_1 c_1 (T - T_1) + m_2 c_2 (T - T_2) + m_3 c_3 (T - T_3) = 0$.Rearranging the terms to solve for $T$:$T(m_1 c_1 + m_2 c_2 + m_3 c_3) = m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3$.Therefore,the final temperature $T$ is given by:$T = \frac{m_1 c_1 T_1 + m_2 c_2 T_2 + m_3 c_3 T_3}{m_1 c_1 + m_2 c_2 + m_3 c_3}$.

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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