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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    The specific heat capacity of a metal at low temperature (T) is given as : \[{{C}_{p}}(kj{{K}^{-1}}k{{g}^{-1}})=32{{\left( \frac{T}{400} \right)}^{3}}\] A 100 gram vessel of this metal is to be cooled from \[20{}^\circ K\] to \[4{}^\circ K\] by a special refrigerator operating at room temperature \[\left( 27{}^\circ C \right)\]. The amount of work required to cool the vessel is :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A)  greater than 0.148 kJ                     

    B)  between 0.148 kJ and 0.028 kJ

    C)  less than 0.028 kJ                        

    D)  equal to 0.002 kJ

    Correct Answer: D

    Solution :

                                    \[Q=\int_{{}}^{{}}{mcdT}\]                 \[=\int\limits_{20}^{4}{0.1\times 32\times }\left( \frac{{{T}^{3}}}{{{400}^{3}}} \right)dT\approx 0.002kJ.\]


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