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.
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To find the rate of combustion of the fuel, we need to calculate the amount of heat transferred from the fuel to the water per minute.
First, we need to find the heat transferred from the fuel to raise the temperature of the water. We can use the formula:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
We are given that the flow rate of water is 3 L/min, which is equivalent to 3000 grams/min (since 1 L of water is equal to 1000 grams). The initial temperature of the water is 27°C, and it is heated to a final temperature of 77°C.
Using the specific heat capacity of water, which is approximately 4.18 J/g°C, we can calculate the heat transferred per minute:
Q = (3000 g/min) * (4.18 J/g°C) * (77°C - 27°C)
Q = 3000 * 4.18 * 50
Q = 627,000 J/min
Now, we need to find the rate of combustion of the fuel. We know that the heat of combustion of the fuel is given as 40,000 J/g.
To find the rate of combustion, we need to divide the heat transferred per minute (Q) by the heat of combustion:
Rate of combustion = Q / Heat of combustion
Rate of combustion = 627,000 J/min / 40,000 J/g
Rate of combustion = 15.675 g/min
Therefore, the rate of combustion of the fuel is 15.675 grams per minute.