202109151148 CYU 1.9 – Physicspupil

How much greater is the rate of heat radiation when a body is at the temperature $40\,^\circ\mathrm{C}$ than when it is at the temperature $20\,^\circ\mathrm{C}$ ?

Background. (Stefan-Boltzmann law of radiation)

$P=\sigma AeT^4$

where $\sigma = 5.67\times 10^{-8}\,\mathrm{J/s}\cdot\mathrm{m^2}\cdot\mathrm{K^4}$ is the Stefan-Boltzmann constant; $A$ is the surface area of the object; and $T$ is its temperature in kelvins.

By temperature conversion formula,

$T_{\mathrm{(K)}}=T_{\mathrm{(^\circ C)}}+273.15$ .

$\begin{aligned} & T_{\mathrm{(^\circ C)}} = 20 \\ \Rightarrow \qquad & T_{\mathrm{(K)}} = 20 + 273.15 = 293.15 \\ & T_{\mathrm{(^\circ C)}} = 40 \\ \Rightarrow \qquad & T_{\mathrm{(K)}} = 40 + 273.15 = 313.15 \\ \end{aligned}$

The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation:

$P=\sigma AeT^4$ .

When the temperature increases from $20\,^\circ\mathrm{C}$ to $40\,^\circ\mathrm{C}$ , ceteris paribus, the percentage change of rate of heat radiation is

$\begin{aligned} & \qquad \frac{P_{\textrm{new}}-P_{\textrm{old}}}{P_{\textrm{old}}} \times 100 \% \\ & = \frac{\sigma AeT_{\textrm{new}}^4-\sigma AeT_{\textrm{old}}^4}{\sigma AeT_{\textrm{old}}^4} \times 100 \% \\ & = \frac{T_{\textrm{new}}^4-T_{\textrm{old}}^4}{T_{\textrm{old}}^4} \times 100 \% \\ & = \frac{313.15^4-293.15^4}{293.15^4}\times 100 \% \\ & \approx 30.2 \% \qquad \textrm{(3\, s.f.)} \\ \end{aligned}$

$\therefore$ The rate of heat transfer increases by about $30\%$ of the original rate.

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