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CMC Medical CMC-Medical VELLORE Solved Paper-2010

  • question_answer
    0.3 kg of hot coffee, which is at \[70{}^\circ C,\] is poured into a cup of mass 0.12 kg. Find the final equilibrium temperature. Take room temperature as \[20{}^\circ C,\]\[{{S}_{coffee}}=4080J\text{/}kg\text{-}K\]and \[{{S}_{cup}}=1020J\text{/}kg\text{-}K.\]

    A)  \[45.5{}^\circ C\]            

    B)  \[55.5{}^\circ C\]

    C)  \[65.5{}^\circ C\]            

    D)  \[40.5{}^\circ C\]

    E)  None of these

    Correct Answer: C

    Solution :

                    Let T be the final equal temperature. Heat lost by coffee = Heat gained by cup \[0.3\times {{s}_{coffee}}\times (70-T)=0.12\times {{s}_{cup}}\times (T-20)\] \[\Rightarrow \]\[0.3\times 4080\,(70-T)=0.12\times 1020\times (T-20)\] \[\Rightarrow \]               \[4\times 70-4T=0.4T-8\] \[\Rightarrow \]                               \[T=\frac{288}{4.4}\]                                 \[=65.5{}^\circ C\]


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