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

  • question_answer
    A body cools from \[50.0{}^\circ C\] to \[49.9{}^\circ C\] in 5s. How long will it take to cool from \[40.0{}^\circ C\]to\[39.9{}^\circ C\]? Assume the temperature of surroundings to be \[30.0{}^\circ C\] and Newton's law of cooling to be valid

    A) 2.5 s                            

    B) 10 s

    C) 20 s                 

    D)        5 s

    Correct Answer: B

    Solution :

    [b] \[\frac{50-49.9}{t}=K\left( \frac{50+49.9}{2}-30 \right)\]         ....(i) \[\frac{40-39.9}{t}=K\left[ \frac{40+39.9}{2}-30 \right]\]  ?(ii) From equations (i) and (ii), we get\[\operatorname{t}\approx 10 s\].


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