Fundamentals of Materials Science and Engineering, Binder Ready Version: An Integrated Approach
Fundamentals of Materials Science and Engineering, Binder Ready Version: An Integrated Approach
5th Edition
ISBN: 9781119175483
Author: William D. Callister Jr., David G. Rethwisch
Publisher: WILEY
Question
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Chapter 17.5, Problem 1QP
To determine

To estimate:

The energy required to raise the temperature of 5 kg (11.0 lbm) for the materials aluminum, brass, aluminum oxide and polypropylene.

Expert Solution & Answer
Check Mark

Answer to Problem 1QP

The energy E value for aluminum is 5.85×105J .

The energy E value for brass is 2.44×105J .

The energy E value for aluminum oxide is 5.04×105J .

The energy E value for polypropylene is 1.25×106J .

Explanation of Solution

Given:

The temperature of materials ranges from 20°C to 150°C (68°F to 300°F) .

The mass of the materials, m=5kg .

The maximum temperature of materials , T2=150°C .

The minimum temperature materials, T1=20°C  .

Explanation:

Write the expression to determine the energy required to raise the temperature of given materials.

E = cpmΔT (I).

Here, the energy is E , and specific heat or heat capacity at constant temperature is cp .

Write the expression to calculate the change in temperature.

ΔT=T2T1 (II).

From Table 19.1, “thermal properties of materials”, write the properties of specific heats as follows:

Aluminum cp = 900 J/kg-K

Brass cp= 375 J/kg-K

Aluminum oxide cp= 775 J/kg-K

Polypropylene cp= 1925 J/kg-K

Conclusion:

Substitute 150°C for T2 and 20°C  for T1 in Equation (II).

ΔT=T2T1=(150+273)K(20+273)K=150+27320273=130K

The energy value for Aluminum.

Substitute 5kg for m , 130K for ΔT and 900J/Kg-K for cp in Equation(I).

E = cpmΔT=( 900 J/kg-K)(5kg)(130K)=5.85×105J

Thus, the energy E value for aluminum is 5.85×105J .

The energy value for Brass.

Substitute 5kg for m , 130K for ΔT and  375 J/kg-K for cp in Equation(I).

E = cpmΔT=( 375 J/kg-K)(5kg)(130K)=2.44×105J

Thus, the energy E value for brass is 2.44×105J .

The energy value for Aluminum oxide .

Substitute 5kg for m , 130K for ΔT and 775 J/kg-K for cp in Equation(I).

E = cpmΔT=( 775 J/kg-K)(5kg)(130K)=5.04×105J

Thus, the energy E value for aluminum oxide is 5.04×105J .

The energy value for Polypropylene.

Substitute 5kg for m , 130K for ΔT and 1925 J/kg-K for cp in Equation(I).

E = cpmΔT=(1925 J/kg-K)(5kg)(130K)=1.25×106J

Thus, the energy E value for polypropylene is 1.25×106J .

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