5. Consider two identical containers ofnegligible mass each containing the same

Question

5. Consider two identical containers ofnegligible mass each containing the same amount of the same liquid. The liquid in container B is cooled to a temperature of -35 oC, the liquid in container A is heated to 95 oC. Half of the liquid from B is then poured into A. After the temperature in A stabilizes, liquid is poured from A back into B until the mass of liquid inthetwo containers is again the same. a. Find the final temperature of the liquid in A. b. Find the final temperature of the liquid in B Assume now that there are two different liquids initially in A and B, but that they have the same density. When the experiment described above is repeated, the final temperature of the liquid remaining in A is found to be 44 oC. (Assume that the two liquids mix uniformly when they are poured together.) c. Find the ratio of the specific heat of the liquid originally in B to the specific heat ofthe liquid originally in A. d. Find the final temperature of.the liquid in B

Explanation / Answer

Q = m x cp x DT Liquid B = Liquid 1. T1 = -35 °C Liquid A = Liquid 2 = T2 = 95°C;

m1 = m2 = m

Q1 = Q2

Q1 = 1/2m1 x cp x (t1 - tf)

Q2 = m1 x cp x (tf - t2)

m1 x cp x (t1 - tf) / 2 = m1 x cp x (tf - t2)

t1 - tf = 2(tf - t2)

-35 - tf = 2tf - 95

95 - 35 = 3tf --------> tf = 60/3 = 20 °C This is the final temperature in liquid A.

b) The same procedure of before but changing the innitial temperature of A to 20 °C

1/2m1 x cp x (tf - t1) = 1/2m1 x cp x (t2 - tf)

tf - t1 = t2 - tf

2tf = t2 + t1

2tf = 20 - 35

tf = -7.5 °C ----> This would be the final temperature in liquid B.

c) Assuming they both have the same density, and the same mass, and we do know the final temperature in A, then:

m1 x cp1 x (-35 - 44) = 2m1 x cp2 x (44 - 95)

cp1 x (-79) = 2Cp2(-51)

cp1 = (2 x -51 / -79) x cp2

cp1 = 1.29cp2 ------> Cp1/Cp2 = 1.29

d) 1/2m1 x cp1 x (tf - t1) = 1/2m1 x cp2 x (t2 - tf)

1/2m1 x 1.29cp2 x (tf - t1) = 1/2m1 x cp2 x (t2 - tf)

1.29(tf - t1) = (t2 - tf)

1.29tf - 1.29t1 = t2 - tf

2.29tf = t2 + 1.29t1

2.29tf = 44 - 1.29(-35)

2.29tf = -1.15

tf = -0.5 °C

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