Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    \[0.1\,{{m}^{3}}\] of water at \[80{}^\circ C\] is mixed with \[0.3\,{{m}^{3}}\] of water at \[60{}^\circ C\]. The final temperature of the mixture is

    A) \[65{}^\circ C\]                                 

    B) \[70{}^\circ C\]

    C) \[60{}^\circ C\]                 

    D)        \[75{}^\circ C\]

    Correct Answer: A

    Solution :

                    Let the final temperature of the mixture be\[t\] Heat lost by water at \[{{80}^{o}}C\]                 \[=ms\Delta t\]                 \[=0.1\times {{10}^{3}}\times {{s}_{water}}\times ({{80}^{o}}-t)\]                 \[(\because \,\,m=V\times d=0.1\times {{10}^{3}}kg)\] Heat gained by water at \[{{60}^{o}}C\]                 \[=0.3\times {{10}^{3}}\times {{s}_{water}}\times (t-{{60}^{o}})\] According to principle of calorimetry Heat lost = Heat gained \[\therefore \]  \[0.1\times {{10}^{3}}\times {{s}_{water}}\times ({{80}^{o}}-t)\]                 \[=0.3\times {{10}^{3}}\times {{s}_{water}}\times (t-{{60}^{o}})\] or            \[({{80}^{o}}-t)=3\times (t-{{60}^{o}})\] or            \[4t={{260}^{o}}\] or            \[t={{65}^{o}}C\]


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