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

  • question_answer
    4 gms of steam at \[100{}^\circ C\] is added to 20 gms of water at \[46{}^\circ C\] in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation \[=\text{ }540\text{ }cal/gm\]. Specific heat of water \[=\text{ }1\text{ }cal/gm-{}^\circ C\].

    A) 18gm                           

    B) 20gm

    C) 22gm

    D) 24gm

    Correct Answer: C

    Solution :

    Heat released by steam inconversion to water at \[100{}^\circ C\] is \[{{Q}_{1}}=\text{ }mL\text{ }=\text{ }4\text{ }x\text{ }540\text{ }=\text{ }2160\text{ }cal.\]Heat required to raise temperature of water from is\[{{Q}_{2}}=\text{mS}\Delta \text{ }\!\!\theta\!\!\text{ =20}\times \text{1}\times \text{54=1080J}\]\[{{Q}_{1}}>{{Q}_{2}}\,\text{and}\,\frac{{{Q}_{1}}}{{{Q}_{2}}}=2\]Hence all steam is not converted to water only half steam shall be converted to water\[\therefore \] Final mass of water \[=20\text{ }+\text{ }2\text{ }=22\text{ }gm\]


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