Answer
$q= 4.20 \times 10^{6} \frac{W}{m^{2}}$
Work Step by Step
Given:
brass 7.5 mm (0.30 in.) thick
Temperatures at faces are 150°C and 50°C (302°F and 122°F)
Assume steady-state heat flow
Required:
heat flux
Solution:
Using Equation 19.5:
$q = - k \frac{ΔT}{Δx}$
For gold, $k = 315 \frac{W}{m.K}$ from Table 19.1.
Substituting the given values:
$q = - k \frac{ΔT}{Δx} = -(315 \frac{W}{m.K})[\frac{(50 +273 K) - (150 + 273 K)}{7.5 \times 10^{-3} m}] = 4.20 \times 10^{6} \frac{W}{m^{2}}$
