Solution:

To find the ratio of the specific heats of two liquids A and B when they are mixed, we can use the principle of conservation of energy, which states that the heat lost by the hotter liquid equals the heat gained by the colder liquid.

 Given:
- Temperature of liquid A, \( T_A = 32°C \)
- Temperature of liquid B, \( T_B = 24°C \)
- Final temperature of the mixture, \( T_f = 28°C \)

Let \( c_A \) and \( c_B \) be the specific heats of liquids A and B, respectively. Since equal masses of both liquids are mixed, we can denote the mass as \( m \).

 Heat Lost by A:
\[
Q_A = m \cdot c_A \cdot (T_A - T_f) = m \cdot c_A \cdot (32 - 28) = 4m c_A
\]

 Heat Gained by B:
\[
Q_B = m \cdot c_B \cdot (T_f - T_B) = m \cdot c_B \cdot (28 - 24) = 4m c_B
\]

 Setting Heat Lost Equal to Heat Gained:
\[
Q_A = Q_B
\]
\[
4m c_A = 4m c_B
\]

Canceling \( 4m \) from both sides:
\[
c_A = c_B
\]

Conclusion:
The ratio of the specific heats of liquids A and B is:
\[
\frac{c_A}{c_B} = 1
\]

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