Two liquids of equal volume are thoroughly mixed. If their specific heat are \(c_{1}, c_{2}\), temperatures \(\theta_{1}, \theta_{2}\) and densities \(\mathrm{d}_{1}, \mathrm{~d}_{2}\) respectively. What is the final temperature of the mixture ? (A) $\left\\{\left(\mathrm{d}_{1} \mathrm{c}_{1} \theta_{1}+\mathrm{d}_{2} \mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \theta_{1}+\mathrm{d}_{2} \theta_{2}\right)\right\\}$ (B) $\left\\{\left(\mathrm{c}_{1} \theta_{1}+\mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \mathrm{c}_{1}+\mathrm{d}_{2} \mathrm{c}_{2}\right)\right\\}$ (C) $\left\\{\left(\mathrm{d}_{1} \theta_{1}+\mathrm{d}_{2} \theta_{2}\right) /\left(\mathrm{c}_{1} \theta_{1}+\mathrm{c}_{2} \theta_{2}\right)\right\\}$ (D) $\left\\{\left(\mathrm{d}_{1} \mathrm{c}_{1} \theta_{1}+\mathrm{d}_{2} \mathrm{c}_{2} \theta_{2}\right) /\left(\mathrm{d}_{1} \mathrm{c}_{1}+\mathrm{d}_{2} \mathrm{c}_{2}\right)\right\\}$

Step by step solution

01

Determine the heat gained and lost by each liquid

First, let's determine the heat gained by liquid 1 (Q1) and the heat lost by liquid 2 (Q2). We'll assume a positive Q represents heat gained and a negative Q represents heat lost. Since the liquids have equal volume, we can say that: Q1 = mass1 × specific heat1 × (Tf - T1) Q2 = mass2 × specific heat UIScreen2FFinHLOJQU-BB]]></

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