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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    A metal ball of mass 0.1 kg is heated upto \[500{}^\circ C\] and dropped into a vessel of heat capacity\[800\,\,J{{K}^{-1}}\] and containing 0.5 kg water. The initial temperature of water an d vessel is \[30{}^\circ C\]. What is the approximate percentage increment in the temperature of the water? [Specific heat capacities of water and metal are, respectively, 4200 \[Jk{{g}^{-1}}\,{{K}^{-1}}\]and 400 \[J\,k{{g}^{-1}}\,{{K}^{-1}}\]] [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) 15%                             

    B) 30%  

    C) 25%                 

    D)   20%

    Correct Answer: D

    Solution :

    Heat lost by metal ball = Heat gained by container and water \[{{m}_{b}}{{s}_{b}}(500-\theta )={{m}_{c}}{{s}_{c}}(\theta -30)+{{m}_{w}}{{s}_{w}}(\theta -30)\] \[0.1\times 400(500-\theta )=800(\theta -30)+0.5\times 4200(\theta -30)\]\[40(500-\theta )=2900(\theta -30)\] \[20000-40\theta =2900\theta -87000\] \[\theta =\frac{107000}{2940}={{36.4}^{o}}C\] % increase in the temperature of water \[=\frac{\Delta \theta }{\theta }\times 100=\frac{36.4-30}{30}\times 100=21.33%\]


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