If three different liquid of different masses specific heats and temperature are mixed with each other then what is the temperature of the mixture at thermal equilibrium. If,
$m_{1}, s_{1}, T_{1} \to$ specification for liquid
$m_{2}, s_{2}, T_{2} \to$ specification for liquid
$m_{3}, s_{3}, T_{3} \to$ specification for liquid

\frac{m_{1} T_{1} + m_{2} T_{2} + m_{3} T_{3}}{m_{1} s_{1} + m_{2} s_{2} + m_{3} s_{3}}

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(m_{1} s_{1} T_{1} + m_{2} s_{2} T_{2} + m_{3} s_{3} T_{3})

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\frac{m_{1} s_{1} T_{1} + m_{2} s_{2} T_{2} + m_{3} s_{3} T_{3}}{m_{1} s_{1} + m_{2} s_{2} + m_{3} s_{3}}

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\frac{s_{1} T_{1} + s_{2} T_{2} + s_{3} T_{3}}{m_{1} s_{1} + m_{2} s_{2} + m_{3} s_{3}}

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Solution

The correct option is D $\frac{m_{1} s_{1} T_{1} + m_{2} s_{2} T_{2} + m_{3} s_{3} T_{3}}{m_{1} s_{1} + m_{2} s_{2} + m_{3} s_{3}}$
Total heat loss is zero.
$⟹ Δ Q = 0$
So, $m_{1} s_{1} (T - T_{1}) + m_{2} s_{2} (T - T_{2}) + m_{3} s_{3} (T - T_{3}) = 0$
$\Rightarrow T = \frac{m_{1} s_{1} T_{1} + m_{2} s_{2} T_{2} + m_{3} s_{3} T_{3}}{m_{1} s_{1} + m_{2} s_{2} + m_{3} s_{3}}$

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Similar questions

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:

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 $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is