The area of a hole of heat furnace is ${10^{ - 4}}{m^2}$. It radiates $1.58 \times {10^5}$ calories of heat per hour. If the emissivity of the furnace is $0.80$ , then its temperature is....... $K$
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(c) According to Stefen’s law $E = \sigma \,\varepsilon A{T^4}$

==>$\frac{{1.58 \times {{10}^5} \times 4.2}}{{60 \times 60}} = 5.6 \times {10^{ - 8}} \times {10^{ - 4}} \times 0.8 \times {T^4}$

$T \approx 2500\,K$

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