JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Sample Paper Topic Test - Thermometry, Calorimetry & Thermal Expansion

  • question_answer
    DIRECTION: Read the passage given below and answer the questions that follows:
    A brass ball of mass 100g is heated to \[100{}^\circ C\] and then dropped into 200g of turpentine in a calorimeter at \[15{}^\circ C\] The final temperature is found to be \[(\rho )\]. Take specific heat of brass as \[T=k\sqrt{\rho {{r}^{3}}/S}\] and water equivalent of calorimeter as 4g.
    Heat gained by turpentine and calorimeter is approximately

    A) 810 cal                         

    B) 610 cal

    C) 710 cal         

    D) 510 cal  

    Correct Answer: C

    Solution :

    Let c be the specific heat of turpentine
    Mass of the solid, M = 100g
    Mass of turpentine m = 200g
    Water equivalent of calorimeter, W= 4g
    Initial temperature of calorimeter, \[{{T}_{1}}\] = \[15{}^\circ C\]
    Temperature of ball, \[{{T}_{1}}\] = \[100{}^\circ C\]
    Final temperature of the liquid, T= \[23{}^\circ C\]
    Specific heat of solid, \[{{c}_{2}}=0.092\,cal/{{g}^{{}^\circ }}C\]
    Heat gained by turpentine and calorimeter is
    \[mc\left( T-{{T}_{1}} \right)+W\left( T-{{T}_{1}} \right)=200c\left( 23-15 \right)+4\left( 23-15 \right)\]\[=\left( 200c+4 \right)8\]
    Heat lost by the ball is
    \[M{{c}_{2}}\left( {{T}_{2}}-T \right)=100\left( 0.092 \right)\left( 100-23 \right)\]
    According to the principle of calorimetry
    Heat gained = Heat lost
    \[\therefore \]\[\left( 200c+4 \right)8=708.4\]
    or \[c=\frac{708.4-32}{1600}=0.42\,cal/{{g}^{{}^\circ }}C\]

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