Question

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?

Answer 1

Rate of flow of water ( V) = 3L/min
= 3 × 10^-3 m³/min
Density of water ( d) = 10³
mass of water flowing per minute = rate of flowing of water × density of water
= 3 × 10^-3 × 10³ = 3 kg/min
temperature change( ∆T) = 77-27= 50°C
specific heat of water ( S) = 4.2 J/g-°C
= 4.2 × 10³ J/Kg.°C
Heat taken by water = mS∆T
= 3 × 4.2 × 10³ × 50
= 63 × 10⁴ J/min
Now,
Let m kg of fuel is utilized per minute .
Heat of combustion of fuel = 4 × 10⁴ J/g = 4 × 10⁴ × 10³ J/kg
So, heat produced = m × 4 × 10^7 J/min

We know,
Heat loss = heat gain
63 × 10⁴ = m × 4 × 10^7
m = 63 × 10^-3/4
= 15.75 × 10^-3 kg/min
= 15.75 g/min

Answer 2

Here, volume of water heated = 3.0 L/min
Mass of water heated, m = 3000 g/min
Rise in temperature, ∆T = 77 - 27 = 50°C
Specific heat of water, C = 4.2 Jg−1 C−1
Amount of heat used,
∆Q = mc∆T = 3000 x 4.2 x 50 = 63 x 104 J / min
Heat of combustion = 4 x 104 J/g
Rate of combustion of fuel = 63 x 104 / 4 x 104 = 15.75 g/min

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