CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

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

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

    B)  \[{{70}^{o}}C\]

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

    D)  \[{{75}^{o}}C\]

    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            \[4\,t={{260}^{o}}\] or            \[t={{65}^{o}}C\]  


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