Question

A container open at the top is made of a perfectly insulating material. A wooden lid of thickness $5\times {10}^{-3}m$ closes the top tightly. Heated oil at temperature $T$ is flowing continuously through the container as shown in the figure. The outer surface of the wooden lid attains a constant temperature of ${127}^{\circ }C$ when the surrounding are at ${27}^{\circ }C$. Calculate
(a) the radiation loss (in $J{s}^{-1}{m}^{-2})$ from the lid and
(b) the temperature $T$ (in ${}^{\circ }C)$ of the oil. The thermal conductivity and emissivity of wood are $0.149W{m}^{-1}{C}^{-1}$ and $0.6$ respectively. The value of the Stefan's constant is $\sigma =\frac{17}{3}\times {10}^{-8}W{m}^{-2}{K}^{-4}$ Neglect heat loss by convection and give answer correct to the nearest whole number).

Answer 1

Let $T_1$ be the temperature of the outer surface of the lid and $T_2$ be the temperature of the surroundings. Then the radiation loss is given by
$\frac{dQ}{dt}=\sigma A\rho [T_1^4-T_2^4]$
$\Rightarrow \frac{1}{A}\frac{dQ}{dt}=e\sigma (T_1^4-T_2^4]=595J{s}^{-1}{m}^{-2}$
The radiation loss is compensated by the flow of heat from the oil to the outer surface of the lid.
If $△T$ is the temperature difference between the outer and inner surface of the lid,
$\frac{dQ}{dt}=\frac{K△TA}{d}$
where $d$ is the thickness of the lid. This implies that $△T={20}^{\circ }$ and the temperature of the oil $T=T_1+△T=127+20={147}^{\circ }C$,