## Materials Science and Engineering: An Introduction

$\frac{dQ}{dt}= 1.44 \times 10^{9} J/h$
Given: brass with 15 mm thickness area = $0.5 m^{2} (5.4 ft^{2})$ $k = 120 \frac{W}{m.K}$ from Table 19.1 Temperatures are 150°C and 50°C Required: heat loss per hour Solution: Let $\frac{dQ}{dt} =- kAt \frac{ΔT}{Δx}$, $\frac{dQ}{dt} = - (120 \frac{J}{ s.m^{2}})(0.5 m^{2})(60 s/min)(60min/hr) [(\frac{(50 + 273 K) -(150+273 K)}{15 \times 10^{-3} m}] = 1.44 \times 10^{9} J/h$