Answer
a. $E_{aluminum} = 5.85 \times 10^{5} J$
b. $E_{brass} = 2.44 \times 10^{5} J$
c. $E_{alumina} = 5.04 \times 10^{5} J$
d. $E_{polypropylene} = 1.25 \times 10^{6} J$
Work Step by Step
Given:
m= 5 kg (11.0 lbm)
Temperature 1 = 20°C (68°F)
Temperature 2 = 150°C (300°F)
a. Aluminum
b. Brass
c. Alumina
d. Polypropylene
Required:
energy required to raise temperature (E)
Solution:
Using the equation:
$E = c_{p}mΔT$
where E is the energy required, $c_{p}$ is the specific heat of material, and ΔT is the temperature change.
From Table 19.1, $c_{p}$ of the given materials is shown:
a. $c_{p (aluminum)} = 900 \frac{J}{kg.K}$
b. $c_{p (brass)} = 375 \frac{J}{kg.K}$
c. $c_{p (alumina)} = 775 \frac{J}{kg.K}$
d. $c_{p (polypropylene)} = 1925 \frac{J}{kg.K}$
Solving for the ΔT,
$ΔT = (150°C - 20°C) = 130 °C = 130 K$*
Note, degrees C and degrees K are only the same for a $\underline{change}$ in temperature.
Substituting the given values:
a. $E_{aluminum} = (900 \frac{J}{kg.K}) (5 kg) (130 K) = 5.85 \times 10^{5} J$
b. $E_{brass} = (375 \frac{J}{kg.K}) (5 kg) (130 K) = 2.44 \times 10^{5} J$
c. $E_{alumina} = (775 \frac{J}{kg.K}) (5 kg) (130 K) = 5.04 \times 10^{5} J$
d. $E_{polypropylene} = (1925 \frac{J}{kg.K}) (5 kg) (130 K) = 1.25 \times 10^{6} J$
