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physics(heat)

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A geyser heats water flowing at the rate of 3L/min from 27 deg C to 77 deg C. if the geyser operates on a burner, what is the rate of combustion of the fuel if its heat of combustion is 40000J/g?

  • physics(heat) - ,

    V/t =3L/min => m/t = 3 kg /min =3/60 =0.05 kg/s
    ΔT = 77-27 = 50ºC
    Specific heat of water c =4 180 J/kg•ºC.
    Amount of heat used per second is
    ΔQ/t =m•c• ΔT/t = 0.05•4180•50 =10450 J/s
    Heat of combustion r = 4•10^7 J/kg

    Rate of combustion of fuel is
    10450/ 4•10^7 =2.61•10^-4 kg/s =
    =0.0157 kg/min =15.7 g/min.

  • physics(heat) - ,

    thanks :)

  • physics(heat) - ,

    thank you so much :)

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