A liquid of mass m and specific heat S is heated to a temperature T. Another liquid of mass $m / 2$ and specific heat $2 S$ is heated to a temperature $2 T$ . If these two liquids are mixed, the resultant temperature of the mixture.

\frac{2}{3}

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\frac{8}{5}

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\frac{3}{3}

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\frac{3}{2}

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Solution

The correct option is D $\frac{3}{2}$ T
Given: A liquid of mass m and specific heat S is heated to a temperature T. Another liquid of mass $\frac{m}{2}$ and specific heat 2S is heated to a temperature 2T. If these two liquids are mixed,
To find the resultant temperature of the mixture.
Solution:
As per given condition,
Specific heat of one body, $s_{1} = S$
Specific heat of another body, $s_{2} = 2 S$
Mass of one body, $m_{1} = m$
Mass of another body, $m_{2} = \frac{m}{2}$
Temperature of one of the body, $T_{1} = T$
Temperature of the other body, $T_{2} = 2 T$
Let the final temperature be $T_{F}$
If the two mixtures are mixed together, then
Heat lost by hot body = Heat gained by cold body
$m_{2} s_{2} (T_{2} - T_{F}) = m_{1} s_{1} (T_{F} - T_{1}) ⟹ \frac{m}{2} \times 2 S \times (2 T - T_{F}) = m \times S \times (T_{F} - T) ⟹ 2 T - T_{F} = T_{F} - T ⟹ 2 T_{F} = 3 T ⟹ T_{F} = \frac{3}{2} T$
is the final temperature of the mixture.

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