NEET Physics Thermometry, Calorimetry & Thermal Expansion NEET PYQ-Thermometry, Calorimetry and Thermal Expansion

  • question_answer
    If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is:           [AIPMT 1999]

    A)        \[{{270}^{o}}C\]       

    B)       \[{{230}^{o}}C\]

    C)  \[{{100}^{o}}C\]       

    D)       \[{{50}^{o}}C\]

    Correct Answer: C

    Solution :

    Heat required by 1 g ice at \[{{0}^{o}}C\] to melt into 1g water at \[{{0}^{o}}C\],
                \[{{Q}_{1}}=mL\]       (L = latent heat of fusion)
                \[=1\times 80=80\text{ cal}\]      \[(L=80\text{ }cal/g)\]
                Heat required by 1 g of water at \[{{0}^{o}}C\] to boil at \[{{100}^{o}}C\],
                \[{{Q}_{2}}=mc\Delta \theta \]  (c = specific heat of water)
                \[=1\times 1\text{ }(100-0)\]        \[(c=1\text{ }cal/g{}^\circ C)\]
                            \[=100\text{ }cal\]
                Thus, total heat required by 1 g of ice to reach a temperature of \[{{100}^{o}}C\],
                Heat available with 1g of steam to condense into 1g of water at \[{{100}^{o}}C\],
                \[Q'=mL'\]         (L' = latent heat of vaporisation)
                \[=1\times 536\text{ }cal\]    (L’ = 536 cal/g)
                = 536 cal
                Obviously, the whole steam will not be condensed and ice will attain temperature of \[{{100}^{o}}C\].
                Thus, the mixture of temperature is \[{{100}^{o}}C\].

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