C

Step 1: When two liquids with different temperatures are mixed, the heat lost by the hotter liquid is equal to the heat gained by the cooler liquid.
Step 2: Let the specific heats of the two liquids be \(c_1\) and \(c_2\), and their masses both be \(m\). The initial temperatures are \(T_1 = 32^{\circ}C\) and \(T_2 = 24^{\circ}C\). The final temperature after mixing is given as \(T_f = 28^{\circ}C\).
Step 3: Using the principle of conservation of energy:
\[ m c_1 (T_1 - T_f) = m c_2 (T_f - T_2) \]
Step 4: We can cancel \(m\) from both sides (since mass is equal) and rearrange the equation:
\[ c_1 (32 - 28) = c_2 (28 - 24) \]
\[ c_1 (4) = c_2 (4) \]
This implies \(c_1 = c_2\).
Therefore, the ratio of the specific heats is \(1:1\) since both are equal.
Therefore, the correct answer is C.