Answer
$\frac{dQ}{dt}= 1.44 \times 10^{9} J/h$
Work Step by Step
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 $\frac{dQ}{dt} =- kAt \frac{ΔT}{Δx}$,
$\frac{dQ}{dt} = - (120 \frac{J}{ s.m^{2}})(0.5 m^{2})(60 s/min)(60min/hr) [(\frac{(50 + 273 K) -(150+273 K)}{15 \times 10^{-3} m}] = 1.44 \times 10^{9} J/h$