Two liquids A and B are at 32°C and 24°C. When mixed in equal masses the temperature of the mixture is found to be 28°C. Their specific heats are in the ratio of

The correct option is 1:1.

Let the mass of liquid A and liquid B be $m$ since they are mixed in equal masses.
Let $c_A$ and $c_B$ be the specific heats of liquids A and B, respectively.
The initial temperature of liquid A is $T_A=32°C$ and the initial temperature of liquid B is $T_B=24°C$.
When mixed, the final equilibrium temperature of the mixture is $T_m=28°C$.

According to the principle of calorimetry, the heat lost by the hotter liquid (liquid A) is equal to the heat gained by the colder liquid (liquid B), assuming no heat loss to the surroundings.

The heat lost by liquid A is given by:
$Q_{\text{lost}}=m\cdot c_A\cdot (T_A-T_m)$

Substituting the given temperatures:
$Q_{\text{lost}}=m\cdot c_A\cdot (32-28)=4m{c}_A$

The heat gained by liquid B is given by:
$Q_{\text{gained}}=m\cdot c_B\cdot (T_m-T_B)$

Substituting the given temperatures:
$Q_{\text{gained}}=m\cdot c_B\cdot (28-24)=4m{c}_B$

Equating the heat lost and the heat gained:
$4m{c}_A=4m{c}_B$

Dividing both sides by $4m$:
$c_A=c_B$

Therefore, the ratio of their specific heats is:
$\frac{c_A}{c_B}=\frac{1}{1}$

Thus, the ratio of their specific heats is 1:1.