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Additional Practice Feb. 12, 2014

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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?

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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

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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 T f be the final temperature of the bucket and T 1 and T 2 be the initial temperatures of Each liter. Then the ΔTs will become ΔT 1 = T f -T 1 and ΔT 2= T f- T 2

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Preferred Solution (cont.) Q 1 = m 1 x C 1 x ΔT 1 and Q 2 = m 2 x C 2 x ΔT 2 Substituting for ΔT 1 and ΔT 2 Q 1 = m 1 x C 1 x (T f -T 1 ) Q 2 = m 2 x C 2 x (T f -T 2 ) Q1 + Q2 =0 m 1 x C 1 x (T f -T 1 ) + m 2 x C 2 x (T f -T 2 ) = 0

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Preferred Solution (cont.) m 1 x C 1 x (T f -T 1 ) + m 2 x C 2 x (T f -T 2 ) = 0 m 1 C 1 T f -m 1 C 1 F 1 +m 2 C 2 T f -m 2 C 2 T 2 =0 m 1 C 1 T f -m 1 C 1 F 1 = -(m 2 C 2 T f -m 2 C 2 T 2 ) m 1 C 1 T f -m 1 C 1 F 1 = m 2 C 2 T 2 -m 2 C 2 T f 1x1x T f - 1x1x T 1 = 1x1x T 2-1x1x T f (Simplify) T f -T 1 =-T f +T 2 Add T f and T 1 to both sides 2T f = T 1 + T 2 Solve for T f Tf = (T1+T2)/2 Only when M1=M2 and C1=C2 Tf= (20 + 40) /2 = 30 °C

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