Question

Find the temperature of an oven if it radiates 8.28cal/s through an opening whose area is 6.1cm2 . Assume that the radiation is close to the black body.​

Answer 1

The temperature of the oven is 1000 K

Explanation:

Given:

Rate of radiation = 8.28 cal/s

Area of the opening = 6.1 cm² = 6.1 × 10⁻⁴ m²

To find out:

Temperature of the oven

Solution:

From Stephan Boltzmann's law we know that

$\boxed{E=\sigma AT^4}$

Where $\sigma=5.67\times 10^{-8}$ W/m²K⁴

Given

$A=6.1\times 10^{-4}$ m²

$E=8.28$ cal/s

$\implies E=8.28\times 4.184$ W

Therefore,

$T^4=\frac{8.28\times 4.184}{5.67\times 10^{-8}\times 6.1\times 10^{-4}}$

$\implies T^4=1\times 10^{12}$

$\implies T=\sqrt[4]{10^{12}}$

$\implies T=10^3$ K

or $T=1000$ K

Hope this answer is helpful.

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