JEE Main & Advanced Physics Transmission of Heat Question Bank Critical Thinking

  • question_answer
    The total energy radiated from a black body source is collected for one minute and is used to heat a quantity of water. The temperature of water is found to increase form \[{{20}^{o}}C\] to \[{{20.5}^{o}}C\]. If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at \[{{20}^{o}}C\], the temperature of water will be [UPSEAT 2004]

    A)            \[{{21}^{o}}C\]                    

    B)            \[{{22}^{o}}C\]

    C)            \[{{24}^{o}}C\]                    

    D)            \[{{28}^{o}}C\]

    Correct Answer: D

    Solution :

                       The total energy radiated from a black body per minute.                    \[Q\propto {{T}^{4}}\] Þ \[\frac{{{Q}_{2}}}{{{Q}_{1}}}={{\left( \frac{2T}{T} \right)}^{4}}=16\] Þ \[{{Q}_{2}}=16{{Q}_{1}}\]                    If m be mass of water taken and S be its specific heat capacity, then \[{{Q}_{1}}=ms(20.5-20)\] and \[{{Q}_{2}}=ms(\theta -20)\]                    \[\theta {}^\circ C=\]Final temperature of water            Þ \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{\theta -20}{0.5}\] Þ \[\frac{16}{1}=\frac{\theta -20}{0.5}\] Þ \[\theta =28{}^\circ C\]


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