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16 December, 16:03

In Joule's experiments, the slow lowering of a weight (through a pulley and cable arrangement) turned a stir - rer in an insulated container of water. As a result of viscosity, the kinetic energy transferred from the stirrer to the water eventually dissipated. In this process the po - tential energy of the weight was first converted to kinetic energy of the stirrer and the water, and then as a result of viscous forces, the kinetic energy of the water was converted to thermal energy apparent as a rise in tem - perature. Assuming no friction in the pulleys and no heat losses, how large a temperature rise would be found in 1 kg of water as a result of a 1-kg weight being lowered 1m?

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  1. 16 December, 16:14
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    The temperature rise will be 0.0023 °C

    Explanation:

    We know that:

    Potential Energy = mgh

    Kinetic Energy = (1/2) mv²

    Thermal Energy = mCΔT

    where,

    m = mass = 1 kg

    h = height = 1 m

    v = velocity

    C = specific heat of water = 4186 J/kg °C

    ΔT = rise in temperature

    According to given situation:

    Potential Energy of Weight = Kinetic Energy of Stirrer and water

    Kinetic energy of stirrer and water = Thermal Energy of water

    Therefore, it implies that:

    Potential Energy of Weight = Thermal Energy of water

    mgh = mCΔT

    ΔT = gh/C

    ΔT = (9.8 m/s²) (1 m) / (4186 J/kg °C)

    ΔT = 0.0023 °C
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