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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Self Evaluation Test - Thermal Properties of Matter

  • question_answer
    A 2 kg copper block is heated to \[500{}^\circ C\]and then it is placed on a large block of ice at \[0{}^\circ C\]. If the specific heat capacity of copper is \[400 J/kg{}^\circ C\]and latent heat of fusion of water is \[3.5\times l{{0}^{5}} J/\] kg, the amount of ice that can melt is

    A) (7/8) kg

    B)        (7/5) kg

    C) (8/7) kg

    D)        (5/7) kg

    Correct Answer: C

    Solution :

    [c] Let x kg of ice can melt Using law of Calorimetry, Heat lost by copper = Heat gained by ice \[2\times 400\times (500-0)=x\,\times 3.5\times 1{{0}^{5}}\] or \[x=\frac{2\times 400\times 500}{3.5\times {{10}^{5}}}=\frac{8}{7}kg\]


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