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KVPY Sample Paper KVPY Stream-SX Model Paper-11

  • question_answer
    A 'thermacole' icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of \[5.0\] cm. If \[4.0\] kg of ice is put in the box, estimate the amount of ice remaining after 6h. The outside temperature is \[45{}^\circ \]C and coefficient of thermal conductivity of thermacole is 0.01 J/s-m-K. (Heat of fusion of water =\[335\times {{10}^{3}}\]J/kg)

    A) \[4.234kg\]                    

    B) \[1.734kg\]

    C) \[3.687kg\]                    

    D) \[0.456kg\]

    Correct Answer: C

    Solution :

    Given, each side of cubical ice box [a] = 30 cm area of 6 faces of box =\[6\times (30\times 30)c{{m}^{2}}\] A = 5400 \[\times \]\[{{10}^{-4}}{{m}^{2}}\] thickness of the box =\[~5.0cm=5.0\times {{10}^{-2}}cm\]
    Mass of ice (m) \[=4.0kg\]
    Time (t) =6h=\[6\times 60\times 60s\]
    Difference in temperature \[\left( \Delta \theta  \right)={{\theta }_{1}}-{{\theta }_{2}}\]\[=45{}^\circ -0=45{}^\circ C\]
    Latent heat of fusion of water (L) \[=335\times {{10}^{3}}j/kg\]
    Coefficient of thermal conductivity (K) \[=0.01J/s-m-K\]
    Let the mass of the ice melted be m? kg. heat supplied  by the surrounding = heat taken by ice in melting
    \[Q=\frac{KA\Delta \theta t}{d}=m'L\] or  \[m'=\frac{KA\Delta \theta t}{Ld}\]
    =\[\frac{0.01\times 5400\times {{10}^{-4}}\times 45\times 6\times 60\times 60}{335\times {{10}^{3}}\times 5\times {{10}^{-2}}}\]=\[0.313\,kg\]
    \[\therefore \]mass remained in the box unmelted \[=m-m'=4-0.313=3.687kg\]


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