   Chapter 11, Problem 66AP

Chapter
Section
Textbook Problem

Three liquids are at temperatures of 10°C, 20°C, and 30°C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 17°C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 28°C. Find the equilibrium temperature when equal masses of the first and third are mixed.

To determine
The equilibrium temperature when equal masses of the first and third are mixed.

Explanation

Given Info: Temperature of three samples 10°C , 20°C , and 30°C respectively. Masses of all the three samples are same, Equilibrium temperature of first and the second sample mixture is 17°C , equilibrium temperature of the second and third sample mixture is 28°C .

Formula to calculate the energy required to increase the temperature is,

Q=mc(TfTi) (1)

• Q is energy required to raise the temperature of a sample,
• m is mass of a sample,
• c is the specific heat a sample,
• Ti is the initial temperature of the brakes,
• Tf is the final temperature of the brakes,

Apply principle of conservation of energy.

Energy lost be second sample is equal to the energy gain in first sample.

Q1=Q2

Use equation (1) to rewrite the above equation for sample 1 and sample 2 mixture.

mc1(TfTi,1)=mc2(TfTi,2)

Substitute 10°C for Ti,1 , 20°C for Ti,2 , 17°C for Tf and simplify.

c1(17°C10°C)=c2(17°C20°C)c2=(73)c1

Apply principle of conservation of energy.

Energy lost be third sample is equal to the energy gain by the second sample.

Q2=Q3 (2)

Use equation (II) to rewrite the above conservation of energy equation for sample 2 and sample 3 mixture.

mc2(TfTi,2)=mc3(TfTi,3)

Substitute 30°C for Ti,3 , 20°C for Ti,2 , 28°C for Tf and simplify.

c2(28°C20°C)=c3(28°C30°C)c3=4c2

Use (73)c1 for c2 in the above expression and rewrite c3 in terms of c1

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