KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • question_answer
    A 10k W drilling machine is used to drill a bore in an aluminium block of mass 8.0 kg. Block is worked on by machine for 2.5 min to drill a hole and 50% of power is used up in heating the aluminium block. Specific heat of aluminium is \[0.91J{{g}^{-1}}{{K}^{-1}}\]. Rise in temperature of block due to drilling will be

    A) \[100{}^\circ C\]

    B) \[103{}^\circ K\]

    C) \[103{}^\circ C\]

    D) \[50{}^\circ C\]

    Correct Answer: C

    Solution :

    Here, power \[=110kW={{10}^{4}}W\], mass \[m=8.0kg=8\times {{10}^{3}}g\],time\[t=2.5\min =2.5\times 60=150s\]and specific heat \[c=0.91J{{g}^{-1}}{}^\circ C\]
    Total energy =power \[\times \]Time =\[{{10}^{4}}\times 150\operatorname{J}=\] As, \[50%\]of energy is lost.
    Hence, thermal energy available,
    \[\Delta Q=\frac{1}{2}\times 15\times {{10}^{5}}=7.5\times {{10}^{5}}J\]
    As, \[\Delta Q=\operatorname{mc}\Delta \operatorname{T}\]\[\Rightarrow \]\[\Delta T=\frac{\Delta Q}{mc}=\frac{7.5\times {{10}^{5}}}{8\times {{10}^{3}}\times 0.91}=103{}^\circ C\]

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