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JEE Main & Advanced Physics Transmission of Heat Question Bank Radiation Stefan's law

  • question_answer
    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

    A)            1500 K                                     

    B)            2000 K

    C)            2500 K                                     

    D)            3000 K

    Correct Answer: C

    Solution :

                       According to Stefen?s law \[E=\sigma \,\varepsilon A{{T}^{4}}\]                    Þ\[\frac{1.58\times {{10}^{5}}\times 4.2}{60\times 60}=5.6\times {{10}^{-8}}\times {{10}^{-4}}\times 0.8\times {{T}^{4}}\]            \[T\approx 2500\,K\]


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