The temperatures of equal masses of three dif | vedclass

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(B) For equal masses $m$,the final temperature $T_{mix}$ of a mixture of two liquids with specific heats $C_1, C_2$ and initial temperatures $T_1, T_2

Vedclass · Accepted Answer

$8: 10: 5$

Vedclass · Answer

(B) For equal masses $m$,the final temperature $T_{mix}$ of a mixture of two liquids with specific heats $C_1, C_2$ and initial temperatures $T_1, T_2$ is given by $T_{mix} = \frac{T_1 C_1 + T_2 C_2}{C_1 + C_2}$.For liquids $A$ and $B$:$20 = \frac{15 C_A + 24 C_B}{C_A + C_B} \Rightarrow 20 C_A + 20 C_B = 15 C_A + 24 C_B \Rightarrow 5 C_A = 4 C_B \Rightarrow C_A = \frac{4}{5} C_B$.For liquids $B$ and $C$:$26 = \frac{24 C_B + 30 C_C}{C_B + C_C} \Rightarrow 26 C_B + 26 C_C = 24 C_B + 30 C_C \Rightarrow 2 C_B = 4 C_C \Rightarrow C_B = 2 C_C$.Expressing all in terms of $C_C$:$C_B = 2 C_C$$C_A = \frac{4}{5} (2 C_C) = \frac{8}{5} C_C$Thus,$C_A : C_B : C_C = \frac{8}{5} C_C : 2 C_C : C_C = 8 : 10 : 5$.

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