A geyser heats water flowing at the rate of 3.0 litres per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10⁴ J/g?

We know here Water is flowing at a rate of 3.0 litre/min.

The geyser heats the water and raises its temperature from 27°C to 77°C. So change in temperature of water or Rise in temperature is

ΔT = 77°C – 27°C = 50°C

We know heat supplied to a substance in raising its temperature is given by relation

ΔQ = mc ΔT

Where ΔQ is the heat supplied to the substance of mass m Kg having specific heat c to raise its temperature by ΔT

We know Specific heat of water is

c = 4.2 J g^-1°C^-1

we will calculate for 1 minute so we will consider mass of water flowing every minute

Now volume of water flowing every minute

V = 3.0 litre/min

We know

Mass = volume × density

Density of water = 1000g/l

So we have mass of water flowing every minute

m = 3.0 litre/min × 1000g/l

i.e. m = 3000 g/min

putting values of m ,c , ΔT in the equation ΔQ = mc ΔT

we get

ΔQ = 3000g/l × 4.2 J g^-1°C^-1 × 50°C

= 6.3 × 10⁵ J/min

So per minute 6.3 × 10⁵ J heat is required by water

But heat of combustion is of gas is

ΔH = 4.0 × 10⁴ J/g

i.e. per gram of gas 4.0 × 10⁴ J heat is released

so the rate of consumption of gas will be

R = ΔQ/ΔH

Where ΔQ is heat consumed per minute and ΔH is heat supplied per gram of gas

So Rate of consumption will be

= 15.75 g/min

That is each minute 15.75 g fuel is being consumed