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    I was wondering if someone could let me know what answer they get for this question so I can find out if I've done it right :rolleyes:

    A power station needs to get rid of energy at a rate of 600MW and does so by warming up a river which flows past it. The river is 30m wide, 3m deep and flows at an average speed of 1.2m/s. How much warmer is the river downstream of the power station?
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    I was wondering if someone could let me know what answer they get for this question so I can find out if I've done it right

    A power station needs to get rid of energy at a rate of 600MW and does so by warming up a river which flows past it. The river is 30m wide, 3m deep and flows at an average speed of 1.2m/s. How much warmer is the river downstream of the power station?
    This is what i'd do...

    First we need the volume of water passing the power station every second (V). This is

    V = 30 \times 3 \times 1.2 = 108 m^3s^{-1}

    Now we need the mass of this body of water. Assuming the standard density of water (1000kg per cubic meter) we can use that

     m = \rho \times V = 1000 \times 108 = 1.08 \times 10^5 kg

    Now we can use the equation

     Q = m c \Delta T

    which is rearranged to give

     \Delta T = \frac{Q}{mc}

    The power lost is 600MW, i.e. 600MJ/s so we can say this is the energy deposited in one seconds worth of water, hence Q = 600MJ. Putting this in and using that c = 4200 J/(kg kelvin) for water we get

     \Delta T = \frac{600 \times 10^6}{1.08 \times 10^5 \times 4200} = 1.3 kelvin

    Hopefully barring mistakes we get the same!
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    (Original post by spread_logic_not_hate)
    This is what i'd do...

    First we need the volume of water passing the power station every second (V). This is

    V = 30 \times 3 \times 1.2 = 108 m^3s^{-1}

    Now we need the mass of this body of water. Assuming the standard density of water (1000kg per cubic meter) we can use that

     m = \rho \times V = 1000 \times 108 = 1.08 \times 10^5 kg

    Now we can use the equation

     Q = m c \Delta T

    which is rearranged to give

     \Delta T = \frac{Q}{mc}

    The power lost is 600MW, i.e. 600MJ/s so we can say this is the energy deposited in one seconds worth of water, hence Q = 600MJ. Putting this in and using that c = 4200 J/(kg kelvin) for water we get

     \Delta T = \frac{600 \times 10^6}{1.08 \times 10^5 \times 4200} = 1.3 kelvin

    Hopefully barring mistakes we get the same!
    Yep I had exactly the same using the same method so I'm obviously doing it right. Thank you! :yep:
 
 
 
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