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 x 10⁴ J/ g?

Answer 1

heya..

here is your 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

It may help you..

Read more on Brainly.in - https://brainly.in/question/1513681#readmore

Answer 2

Hey mate ^_^

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Answer:
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Volume of water heated = 3.0 L/min

Mass of water heated = 3000 g/min

∆T = 77 - 27 = 50°C

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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