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

The correct option is 3T/2.

To find the resulting temperature of the mixture, we can use the principle of calorimetry. In an isolated system, the heat lost by the hotter body is equal to the heat gained by the colder body.

Let the final temperature of the mixture be
$T_f$
.

First, let's write down the parameters for both liquids:
For the first liquid (hotter liquid):
Mass,
$m_1=m$
Specific heat,
$c_1=c$
Initial temperature,
$T_1=2T$

For the second liquid (colder liquid):
Mass,
$m_2=\frac{m}{2}$
Specific heat,
$c_2=2c$
Initial temperature,
$T_2=T$

Since the first liquid is at a higher temperature, it will lose heat, and the second liquid will gain heat.

The heat lost by the first liquid is given by:
$Q_{\text{lost}}=m_1{c}_1(T_1-T_f)$

$Q_{\text{lost}}=mc(2T-T_f)$

The heat gained by the second liquid is given by:
$Q_{\text{gained}}=m_2{c}_2(T_f-T_2)$

$Q_{\text{gained}}=\left(\frac{m}{2}\right)\left(2c\right)(T_f-T)$
Simplifying the constants:

$Q_{\text{gained}}=mc(T_f-T)$

Applying the principle of calorimetry:
$Q_{\text{lost}}=Q_{\text{gained}}$

$mc(2T-T_f)=mc(T_f-T)$

Dividing both sides by
$mc$
:
$2T-T_f=T_f-T$

Rearranging terms to solve for
$T_f$
:
$2T+T=T_f+T_f$

$3T=2T_f$

$T_f=\frac{3T}{2}$

Thus, the resulting temperature of the mixture is
$\frac{3T}{2}$
.