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(D) Let $m$ be the mass of each liquid and $S_A, S_B, S_C$ be their specific heat capacities.For the mixture of $A$ and $B$:$m S_A (15 - 10) = m S_B (

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$16$

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(D) Let $m$ be the mass of each liquid and $S_A, S_B, S_C$ be their specific heat capacities.For the mixture of $A$ and $B$:$m S_A (15 - 10) = m S_B (25 - 15)$$5 S_A = 10 S_B \Rightarrow S_A = 2 S_B$For the mixture of $B$ and $C$:$m S_B (30 - 25) = m S_C (40 - 30)$$5 S_B = 10 S_C \Rightarrow S_B = 2 S_C \Rightarrow S_C = \frac{S_B}{2}$For the mixture of $A$ and $C$ with final temperature $	heta$:$m S_A (	heta - 10) = m S_C (40 - 	heta)$Substitute $S_A = 2 S_B$ and $S_C = \frac{S_B}{2}$:$2 S_B (	heta - 10) = \frac{S_B}{2} (40 - 	heta)$$4 (	heta - 10) = 40 - 	heta$$4	heta - 40 = 40 - 	heta$$5	heta = 80$$	heta = 16^{\circ}C$

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