Radiation Temp
Consider a 5 cm long cubic piece of burning hot coal with a surface emissivity of 0.2 is buried in sand as a campfire is put out. If the radiation emitted by the piece of coal is 1 W, what is the absolute temperature in Celsius?
Expand Hint
For radiation emitted by a body:
$$$\dot{Q}=\varepsilon \sigma AT^4$$$
where
$$\varepsilon$$
is the body’s emissivity,
$$\sigma$$
is the Stefan-Boltzmann constant
$$=5.67 \times 10^{-8}W/(m^2 \cdot K^4)$$
,
$$A$$
is the body’s surface area, and
$$T$$
is the absolute temperature.
Hint 2
Surface area of a cube:
$$$A=6a^2$$$
where
$$a$$
is the cube’s edge.
Surface area of a cube:
$$$A=6a^2$$$
where
$$a$$
is the cube’s edge.
$$$A=6(0.05m)^2=0.015m^2$$$
For radiation emitted by a body:
$$$\dot{Q}=\varepsilon \sigma AT^4$$$
where
$$\varepsilon$$
is the body’s emissivity,
$$\sigma$$
is the Stefan-Boltzmann constant
$$=5.67 \times 10^{-8}W/(m^2 \cdot K^4)$$
,
$$A$$
is the body’s surface area, and
$$T$$
is the absolute temperature.
$$$1W=(0.2)(5.67 \times 10^{-8}\frac{W}{m^2 \cdot K^4})(0.015m^2)T^4$$$
$$$T=\left ( \frac{1}{(0.2)(5.67 \times 10^{-8})(0.015)}K^4\right )^{0.25}=276.9K$$$
Converting to Celsius:
$$$T=276.9K-273.15=3.75^{\circ}C$$$
3.75°C
Time Analysis
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