A liquid of mass m and specific heat c is hea | vedclass

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(D) Let the final temperature of the mixture be $T_{mix}$.According to the principle of calorimetry,the heat lost by the hotter liquid equals the heat

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$(3/2) T$

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(D) Let the final temperature of the mixture be $T_{mix}$.According to the principle of calorimetry,the heat lost by the hotter liquid equals the heat gained by the colder liquid.Heat lost by liquid $1$ $(m, c, 2T)$: $Q_{lost} = m \cdot c \cdot (2T - T_{mix})$.Heat gained by liquid $2$ $(m/2, 2c, T)$: $Q_{gained} = (m/2) \cdot (2c) \cdot (T_{mix} - T)$.Equating the two: $m \cdot c \cdot (2T - T_{mix}) = (m/2) \cdot (2c) \cdot (T_{mix} - T)$.Simplifying the equation: $2T - T_{mix} = T_{mix} - T$.$3T = 2T_{mix}$.Therefore,$T_{mix} = (3/2) T$.

$A$ liquid of mass $m$ and specific heat $c$ is heated to a temperature of $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is heated to a temperature of $T$. If these two liquids are mixed,what will be the final temperature of the mixture?

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