NEET Sample Paper NEET Sample Test Paper-14

  • question_answer A 1.0 g sample of substance A at \[100{}^\circ C\] is added to 100 mL of \[{{\text{H}}_{\text{2}}}\text{O}\]at \[25{}^\circ C\]. Using separate 100 mL portions of\[{{\text{H}}_{\text{2}}}\text{O,}\]the procedure is repeated with substance B and then with substance C. How will the final temperatures of the water compare?
    Substance Specific Heat
    A \[0.60\,J{{g}^{-1}}{{\,}^{o}}{{C}^{-1}}\]
    B \[0.40\,J{{g}^{-1}}{{\,}^{o}}{{C}^{-1}}\]
    C \[0.20\,J{{g}^{-1}}{{\,}^{o}}{{C}^{-1}}\]

    A) \[{{T}_{C}}>{{T}_{B}}>{{T}_{A}}\]            

    B) \[{{T}_{B}}>{{T}_{A}}>{{T}_{C}}\]

    C) \[{{T}_{A}}>{{T}_{B}}>{{T}_{C}}\]            

    D) \[{{T}_{A}}={{T}_{B}}={{T}_{C}}\] 

    Correct Answer: C

    Solution :

    Heat absorbed or evolved, \[\Delta Q=ms\Delta t.\] Where,\[m=\] mass of substance, s = specific heat\[\Delta t=\] temperature difference It means \[s\propto \frac{1}{\Delta t}.\] Higher the temperature of the given solution, lesser is the temperature difference. So higher is the specific heat. As, the specific heat order is A > B > C. So, the order of temperature of solution is A > B > C. Hence, the correct option is [c].

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