2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Thermometry, Calorimetry & Thermal Expansion

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

  • question_answer
    An experiment takes 10 minutes to raise the temperature of water in a container from \[0{}^\circ C\]to \[100{}^\circ C\]and another 55 minutes to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be\[1 cal/g{}^\circ C\], the heat of vapourisation according to this experiment will come out to be :

    A) 560 cal/g

    B) 550 cal/g

    C) 540 cal/g

    D) 530 cal/g

    Correct Answer: B

    Solution :

    [b] As \[\operatorname{Pt}=mC\Delta T\] So, \[\operatorname{P}\times 10\times 60 = mC 100\]       ...(i) and \[\operatorname{P}\times 55\times 60 = mL\]                         ...(ii) Dividing equation (i) by (ii) we get \[(\theta -{{\theta }_{0}})\] \[\therefore  L = 550 cal./g.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner