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(Take C(w)=4180 J kg^(-1) K^(-1)) A bloc...

(Take `C_(w)=4180 J kg^(-1) K^(-1))` A block of weight 5kg falls from a height of 5m into a bucket containing 10L of water. If all the mechanical energy of the block is used to heat water, find the rise in temperature of water. Take `g= 10ms^(-2)`

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Knowledge Check

  • The height of a waterfall is 84 m. Assuming that the entire kinetic energy of falling water is converted into heat, the rise in temperature of the water will be: (g= 10 ms^(-2), J=4.2joule//cal)

    A
    `0.2^(@)C`
    B
    `1.960^(@)C`
    C
    `0.96^(@)C`
    D
    `0.0196^(@)C`

  • In a water-fall the water falls from a height of 100 m . If the entire K.E. of water is converted into heat, the rise in temperature of water will be

    A
    `0.23^(@)C`
    B
    `0.46^(@)C`
    C
    `2.3^(@)C`
    D
    `0.023^(@)C`

  • Of two masses of 5 kg each falling from height of 10 m , by which 2 kg water is stirred. The rise in temperature of water will be

    A
    `2.6^(@)C`
    B
    `1.2^(@)C`
    C
    `0.32^(@)C`
    D
    `0.12^(@)C`

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    Water at Bhakra dam falls through a height of 210 m. Assuming that the whole of the energy due to fall is converted into heat, calculate the rise in temperature of water. Take J = 4.2 J cal^(-1) and g = 9.8 ms^(-2) .

    A man of weight 50kg carries an object to a height of 20m in a time of 10s .The power used by the man in this process is 2000W then find the weight of the object carried by the man take g=10ms^(-2)

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    A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel =460 K//kg^(@)//C,g=10 m//s^(2) )

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