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Additional Practice

Additional Practice. Feb. 12, 2014. 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

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Additional Practice

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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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