Given:

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

The mass of the materials, m=5 kg .

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 = cpm ΔT (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=T2−T1 (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=T2−T1=(150+273)K−(20+273)K=150+273−20−273=130 K

The energy value for Aluminum.

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

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

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

The energy value for Brass.

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

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

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

The energy value for Aluminum oxide .

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

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

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

The energy value for Polypropylene.

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

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

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