Posted by ridhi on .
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?
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.
thank you so much :)