Introduction To Chemistry 5th Edition
Introduction To Chemistry 5th Edition
5th Edition
ISBN: 9781260162097
Author: BAUER
Publisher: MCG
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Chapter 6, Problem 125QP
Interpretation Introduction

Interpretation:

The specific heat of pipe is to be determined.

Concept Introduction:

Specific heat of a substance is defined as the amount of heat required to raise the temperature of 1 g of that substance by 1°C .

Expert Solution & Answer
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Answer to Problem 125QP

Solution:

The specific heat of pipe is found to be 0.450 J/g°C .

Explanation of Solution

Given Information: Mass of pipe and water is 175 g and 100 g , respectively. The temperatures of pipe and water are 78.24°C and 25.00°C , respectively. The final temperature of the mixture is 33.43°C .

Heat transfer takes place from high temperature to low temperature. So, the temperature of the pipe will decrease and that of the water will increase to attain thermal equilibrium. At thermal equilibrium, temperatures of both pipe and water will be equal.

The heat lost by the pipe can be calculated as follows.

qpipe=mpipe×Cpipe×ΔTpipe

Here, qpipe is the heat lost by the pipe, mpipe is the given mass of the pipe, Cpipe is the specific heat of the pipe, and ΔTpipe is the decrease in temperature of the pipe.

Substitute 175 g for mpipe and T°C78.24°C for ΔTCu in the above equation. Here, T°C is the temperature of pipe and water at thermal equilibrium.

qpipe=150 g×Cpipe×T°C78.24°C

Substitute 33.43°C for T°C in the above equation.

qCu=175 g×Cpipe×33.43°C78.24°C=175 g×Cpipe×44.81°C=7841.75 g°CCpipe

The heat gained by the water can be calculated as follows.

qwater=mwater×Cwater×ΔTwater

Here, qwater is the heat gained by the water, mwater is the given mass of the water, Cwater is the specific heat of the water, and ΔTwater is the increase in temperature of the water.

Substitute 100 g for mwater , 4.184 J/g°C for Cwater , and T°C25.00°C for ΔTwater in the above equation. Here, T°C is the temperature of water and pipe at thermal equilibrium.

qwater=100 g×4.184 J/g°C×T°C25.00°C

Substitute 33.43°C for T°C in the above equation.

qwater=100 g×4.184 J/g°C×33.43°C25.00°C=100 g×4.184 J/g°C×8.43°C=3527.12 J

As no heat is lost from the water, according to the law of conservation of energy, the total heat of the system remains conserved.

qwater+qpipe=0

The specific heat of the copper can be calculated using the above equation and the calculated values of qwater and qCu are as follows.

3527.12 J+7841.75 g°CCpipe=07841.75 g°CCpipe=3527.12 JCpipe=3527.12 J7841.75 g°CCpipe=0.450 J/g°C

Therefore, the specific heat of the pipe is 0.450 J/g°C and using table 6.2, it could be said that the pipe is made up of chromium s .

Conclusion

The specific heat of the pipe is 0.450 J/g°C and the material that the pipe is made up of is chromium s .

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Chapter 6 Solutions

Introduction To Chemistry 5th Edition

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