JEE Main & Advanced Physics Transmission of Heat Sample Paper Topic Test - Transmission Of Heat

  • question_answer
    Direction: Consider a spherical body A of radius R which placed concentrically in a hollow enclosure H, of radius 4R as shown in the figure. The temperature of the body A and H are \[{{T}_{A}}\] and \[{{T}_{H}},\] respectively.
    Emissivity, transitivity and reflectivity of two bodies A and H are \[\text{(}{{e}_{A}},\text{ }{{e}_{H}})\text{ (}{{t}_{A}},\text{ }{{t}_{H}}\text{)}\] and \[({{r}_{A}},\text{ }{{r}_{H}})\] respectively.
    For answering following questions assume no absorption of the thermal energy by the space in-between the body and enclosure as well as outside the enclosure and all radiations to be emitted and absorbed normal to the surface.
    [Take \[\sigma \times \,4\pi {{R}^{2}}\,\times \,{{300}^{4}}=\,\beta J{{s}^{-1}}\]]
    Consider two cases, first one in which A is a perfect black body and the second in which A is a non-black body. In both the cases, temperature of body A is same equal to 300K and H is at temperature 600K. For H, \[t=0\] and \[a\ne 1\]. For this situation, mark out the correct statement.

    A) The bodies lose their distinctiveness inside the enclosure and both of them emit the same radiation as that of the black body.

    B) The rate of heat loss by A in both cases is the same and is equal to\[\beta \,J{{s}^{-1}}\].

    C) The rates of heat loss by A in both the cases are different.

    D) From this information we can calculate exact rate of heat loss by A in different cases.

    Correct Answer: C

    Solution :

    [c]        
    If \[\beta =\sigma \times 4\pi {{R}^{2}}\times \,{{300}^{4}},\]
    Then \[\sigma \times \,4\pi \,{{(4R)}^{2}}\times \,{{600}^{4}}\,=256\beta \,=\gamma \]
    Let \[{{a}_{H}}\,={{e}_{H}}=0.5\]
    and for A in 2nd case, \[{{e}_{A}}=\,{{a}_{A}}=\,0.5\]
    For 1st case, \[{{P}_{emitted}}\,=\,\beta \,J{{s}^{-1}}\]
    \[{{P}_{absorbed}}\,=\,\frac{\gamma }{2}\,+\frac{\beta }{2}\]
    Rate at which energy is lost, \[P=\left( \beta -\frac{\gamma }{2}-\frac{\beta }{2} \right)\,J{{s}^{-1}}\]
    For 2nd case,
    \[{{P}_{emitted}}=\left( \frac{\beta }{2}+\,\frac{\beta }{8}+\frac{\beta }{32}+.... \right)\,+\left( \frac{\gamma }{4}+\frac{\gamma }{16}+... \right)\]
    \[=\,\frac{2\beta }{3}+\,\frac{\gamma }{3}\]
    \[{{P}_{absorbed}}=\,\left( \frac{\beta }{8}+\,\frac{\beta }{32}+.... \right)\,+\,\left( \frac{\gamma }{4}+\,\frac{\gamma }{16}+... \right)\,\,=\frac{\beta }{6}+\frac{\gamma }{3}\]
    Rate at which heat is lost, \[P=\,\frac{\beta }{2}\].


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