A liquid of mass $m$ and specific heat $c$ is at a temperature $2 T$ . Another liquid of mass $m / 2$ and specific heat $2 c$ is at a temperature $T$ . If these two liquids are mixed, the resulting temperature of the mixture is:

$(\frac{2}{3}) T$

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$(\frac{8}{5}) T$

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$(\frac{3}{5}) T$

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$(\frac{3}{2}) T$

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Solution

The correct option is D
$(\frac{3}{2}) T$

Given:
Mass of liquid 1, $m_{1} = m$ ,
Specific heat of liquid 1, $c_{1} = c$ ,
Initial temperature of liquid 1, $T_{1} = 2 T$
Mass of liquid 2, $m_{2} = \frac{m}{2}$
Specific heat of liquid 2, $c_{2} = 2 c$ ,
Initial temperature of liquid 2, $T_{2} = T$

Let the temperature of mixture be $T_{m}$ .

Heat lost by liquid 1 = Heat gained by liquid 2.
$m_{1} c_{1} (T_{1} - T_{m}) = m_{2} c_{2} (T_{m} - T_{2})$
$m c (2 T - T_{m}) = \frac{m}{2} \times 2 c \times (T_{m} - T)$
$T_{m} = \frac{3}{2} T$

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