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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Calorimetry

  • question_answer
    A lead bullet of 10 g travelling at 300 m/s strikes against a block of wood and comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (Specific heat of lead = 150J/kg, K)                    [EAMCET 2001]

    A)            100°C                                       

    B)            125°C

    C)            150°C                                       

    D)            200°C

    Correct Answer: C

    Solution :

                       Since specific heat of lead is given in Joules, hence use \[W=Q\] instead of \[W=JQ\].            Þ\[\frac{1}{2}\times \left( \frac{1}{2}m{{v}^{2}} \right)=m.c.\Delta \theta \]Þ\[\Delta \theta =\frac{{{v}^{2}}}{4c}=\frac{{{(300)}^{2}}}{4\times 150}=150{}^\circ C\].


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