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 it's heat of combustion is 4.0 × 10⁴ J/g??

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Answer 1

$\huge\mathbb{Answer:-}$

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



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Answer 2

$\huge\bold{Solution}$

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Given:-

Volume of water flowing = 3.0 L/min

= 3.0 × 10³ cm³ /min

Mass of water flowing, m = Volume of water flowing × Density

m = V × rho

=> 3000 cm³ / min × 1g /cm³

therefore, m = 3000g /min

Rise in temperature, △T = 77°C – 27°C = 50°C

Specific heat of water, C = 4.2 J g-¹ °C-¹

Heat of combustion = 4 × 10⁴ J/g

then, amount of heat used, △Q = mC△T

= 3000g/min × 4.2 J/g°C-¹ × 50°C

= 6.3 × 10^5 J/min

thus ,

Rate of combustion of fuel

=> (6.3 × 10^5 J/min)/(4 × 10⁴ J/g)

(Heat used )/(Heat of Combustion)

=> 15.75g/min ≈ 16g per min.

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