Let \\\\(m\\\\) be the mass of each liquid and \\\\(S_{A}, S_{B}, S_{C}\\\\) be specific heats of liquids \\\\(A, B\\\\) and \\\\(C\\\\) respectively. When \\\\(A\\\\) and \\\\(B\\\\) are mixed. The final temperature is \\\\(16^{\\circ}C\\\\).\\\\(\ herefore\\\\) Heat gained by \\\\(A =\\\\) heat lost by \\\\(B\\\\)i.e., \\\\(mS_{A} = (16 - 12) = mS_{B} (19 - 16)\\\\)i.e., \\\\(S_{B} = \\dfrac {4}{3}S_{A} ..... (i)\\\\)When \\\\(B\\\\) and \\\\(C\\\\) are mixed. Heat gained by \\\\(B =\\\\) Heat lost by \\\\(C\\\\)i.e., \\\\(mS_{B} = (23 - 19) = mS_{C} (28 - 23)\\\\)i.e., \\\\(S_{C} = \\dfrac {4}{5}S_{B} ..... (ii)\\\\)From eq. (i) and (ii)\\\\(S_{C} = \\dfrac {4}{5}\ imes \\dfrac {4}{3} S_{A} = \\dfrac {16}{15}S_{A}\\\\)When \\\\(A\\\\) and \\\\(C\\\\) are mixed, let the final temperature be \\\\(\ heta\\\\)\\\\(\ herefore mS_{A} (\ heta - 12) = mS_{C} (28 - \ heta)\\\\)i.e., \\\\(\ heta - 12 = \\dfrac {16}{15} (28 - \ heta)\\\\)By solving, we get,\\\\(\ heta = \\dfrac {628}{31} = 20.26^{\\circ}C\\\\).