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Calculating Final Temperature of Mixed Liquids at Different Initial Temperatures

This guide explores the process of determining the final temperature when mixing two equal quantities of liquids at different temperatures. Specifically, we investigate a scenario where one liquid is at 20°C and the other at 40°C. The solution begins with a simple average method, yielding 30°C. However, the preferred method employs the principle of specific heat and heat exchange, confirming the result through a thorough mathematical analysis of heat transfer in the system. Ultimately, the final equilibrium temperature remains consistent, emphasizing the importance of heat conservation.

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Calculating Final Temperature of Mixed Liquids at Different Initial Temperatures

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  1. Additional Practice Feb. 12, 2014

  2. Mixing Liquids at Different Temps • Two Equal Quantities (1 liter each) of Liquids are at different Temperatures • One is at 20 °C the other is at 40 °C • What is the final temperature after the two are allowed to come to Thermal Equilibrium in a 2 liter bucket?

  3. Solution #1 “Cheap Short Cut” • Since the two quantities are the same add the two temps and divide by 2 • 20 +40 =60/2 = 30 °C

  4. Preferred Solution • Approach the problem form the standpoint of Specific Heat Q • Q1 +Q2 = 0 since there is no heat added or lost to the system. • Let Tfbe the final temperature of the bucket and T1 and T2 be the initial temperatures of Each liter. • Then the ΔTs will become ΔT1=Tf-T1and ΔT2= Tf-T2

  5. Preferred Solution (cont.) • Q1= m1 x C1 x ΔT1 andQ2= m2x C2x ΔT2 • Substituting for ΔT1 andΔT2 • Q1= m1 x C1 x (Tf-T1 ) • Q2= m2 x C2x (Tf-T2) • Q1 + Q2 =0 • m1 x C1 x (Tf-T1 ) + m2 x C2 x (Tf-T2) = 0

  6. Preferred Solution (cont.) • m1 x C1 x (Tf-T1 ) + m2 x C2 x (Tf-T2) = 0 • m1C1Tf-m1C1F1+m2C2Tf-m2C2T2=0 • m1C1Tf-m1C1F1= -(m2C2Tf-m2C2T2) • m1C1Tf-m1C1F1= m2C2T2-m2C2Tf • 1x1xTf-1x1xT1= 1x1xT2-1x1xTf (Simplify) • Tf-T1=-Tf+T2 Add Tf and T1 to both sides • 2Tf = T1 +T2 Solve for Tf • Tf = (T1+T2)/2 Only when M1=M2 and C1=C2 • Tf= (20 + 40) /2 = 30 °C

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