Chapter 17.5, Problem 1QP

To determine

To estimate:

The energy required to raise the temperature of $\text{5 kg (11}{\text{.0 lb}}_{\text{m}}\text{)}$ for the materials aluminum, brass, aluminum oxide and polypropylene.

Answer to Problem 1QP

The energy $E$ value for aluminum is $5.85\times {10}^5\text{J}$.

The energy $E$ value for brass is $2.44\times {10}^5\text{J}$.

The energy $E$ value for aluminum oxide is $5.04\times {10}^5\text{J}$.

The energy $E$ value for polypropylene is $1.25\times {10}^6\text{J}$.

Explanation of Solution

Given:

The temperature of materials ranges from $\text{20°C to 150°C }\left(\text{68°F to 300°F}\right)$.

The mass of the materials, $m=5\text{kg}$.

The maximum temperature of materials , $T_2=\text{150°C}$.

The minimum temperature materials, $T_1=\text{20°C }$.

Explanation:

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

$E\text{ = }{c}_{p}m\Delta T$ (I).

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

Write the expression to calculate the change in temperature.

$\Delta T=T_2-T_1$ (II).

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

Aluminum ${\text{c}}_p\text{ = 900 J/kg-K}$

Brass ${\text{c}}_p\text{= 375 J/kg-K}$

Aluminum oxide $c_p\text{= 775 J/kg-K}$

Polypropylene $c_p\text{= 1925 J/kg-K}$

Conclusion:

Substitute $\text{150°C}$ for $T_2$ and $\text{20°C }$ for $T_1$ in Equation (II).

$\begin{array}{c}\Delta T=T_2-T_1\\ =\left(\text{150+273}\right)\text{K}-\left(\text{20+273}\right)\text{K}\\ \text{=150+273}-20-273\\ =130\text{K}\end{array}$

The energy value for Aluminum.

Substitute $5\text{kg}$ for $m$ , $130\text{K}$ for $\Delta T$ and $900\text{J/Kg-K}$ for ${\text{c}}_p$ in Equation(I).

$\begin{array}{c}E\text{ = }{c}_{p}m\Delta T\\ =\left(\text{ 900 J/kg-K}\right)\left(5\text{kg}\right)\left(130\text{K}\right)\\ =5.85\times {10}^5\text{J}\end{array}$

Thus, the energy $E$ value for aluminum is $5.85\times {10}^5\text{J}$.

The energy value for Brass.

Substitute $5\text{kg}$ for $m$ , $130\text{K}$ for $\Delta T$ and $\text{ 375 J/kg-K}$ for ${\text{c}}_p$ in Equation(I).

$\begin{array}{c}E\text{ = }{c}_{p}m\Delta T\\ =\left(\text{ 375 J/kg-K}\right)\left(5\text{kg}\right)\left(130\text{K}\right)\\ =2.44\times {10}^5\text{J}\end{array}$

Thus, the energy $E$ value for brass is $2.44\times {10}^5\text{J}$.

The energy value for Aluminum oxide .

Substitute $5\text{kg}$ for $m$ , $130\text{K}$ for $\Delta T$ and $\text{775 J/kg-K}$ for ${\text{c}}_p$ in Equation(I).

$\begin{array}{c}E\text{ = }{c}_{p}m\Delta T\\ =\left(\text{ 775 J/kg-K}\right)\left(5\text{kg}\right)\left(130\text{K}\right)\\ =5.04\times {10}^5\text{J}\end{array}$

Thus, the energy $E$ value for aluminum oxide is $5.04\times {10}^5\text{J}$.

The energy value for Polypropylene.

Substitute $5\text{kg}$ for $m$ , $130\text{K}$ for $\Delta T$ and $\text{1925 J/kg-K}$ for ${\text{c}}_p$ in Equation(I).

$\begin{array}{c}E\text{ = }{c}_{p}m\Delta T\\ =\left(\text{1925 J/kg-K}\right)\left(5\text{kg}\right)\left(130\text{K}\right)\\ =1.25\times {10}^6\text{J}\end{array}$

Thus, the energy $E$ value for polypropylene is $1.25\times {10}^6\text{J}$.