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Two identical bodies have temperatures $277^{\circ} \mathrm{C}$ and $67^{\circ} \mathrm{C}$. If the surroundings temperature is $27^{\circ} \mathrm{C}$, the ratio of loss of heats of the two bodies during the same interval of time is (approximately)
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Verified Answer
The correct answer is:
19:1
$\begin{aligned}
& T_1=277^{\circ}=277+273=550 \mathrm{~K} \\
& T_2=67^{\circ} \mathrm{C}=67+273=340 \mathrm{~K}
\end{aligned}$
Temperature of surrounding,
$T=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\begin{aligned}
\text { Ratio of loss of heat } & =\frac{T_1^4-T^4}{T_2^4-T^4} \\
& =\frac{\left(\frac{T_1}{T_2}\right)^4-1}{\left(\frac{T_2}{T_1}\right)^4-1}=\frac{\left(\frac{550}{300}\right)^4-1}{\left(\frac{340}{300}\right)^4-1} \\
& =\frac{9.5}{0.5}=\frac{19}{1}
\end{aligned}$
& T_1=277^{\circ}=277+273=550 \mathrm{~K} \\
& T_2=67^{\circ} \mathrm{C}=67+273=340 \mathrm{~K}
\end{aligned}$
Temperature of surrounding,
$T=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\begin{aligned}
\text { Ratio of loss of heat } & =\frac{T_1^4-T^4}{T_2^4-T^4} \\
& =\frac{\left(\frac{T_1}{T_2}\right)^4-1}{\left(\frac{T_2}{T_1}\right)^4-1}=\frac{\left(\frac{550}{300}\right)^4-1}{\left(\frac{340}{300}\right)^4-1} \\
& =\frac{9.5}{0.5}=\frac{19}{1}
\end{aligned}$
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