Materials Science and Engineering: An Introduction

$q= 4.20 \times 10^{6} \frac{W}{m^{2}}$
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: $q = - k \frac{ΔT}{Δx} = -(315 \frac{W}{m.K})[\frac{(50 +273 K) - (150 + 273 K)}{7.5 \times 10^{-3} m}] = 4.20 \times 10^{6} \frac{W}{m^{2}}$