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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 - CBSE (Science) Class 11 - Physics

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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?

Solution 1

Volume of water heated = 3.0 litre per minute Mass of water heated, m = 3000 g per minute Increase in temperature,

`triangle T = 77^@C - 27^@C =50^@C`

Specific heat of water, `c = 4.2 Jg^(-1) ""^@ C^(-1)`

Amount of heat used =  `Q = mctriangleT`

or `Q = 3000 "gmin"^(-1) xx 4.2 Jg^(-1) ""^@C^(-1)xx50^@C`

`= 63 xx 10^4 J min^(-1)`

Rate of combustion of fuel = `(63xx10^4 J min^(-1))/(4.0xx10^4 Jg^(-1))` = 15.75 `"g min"^(-1)`

 

Solution 2

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

The geyser heats the water, raising the temperature from 27°C to 77°C.

Initial temperature, T1 = 27°C

Final temperature, T2 = 77°C

∴Rise in temperature, ΔT = T2 – T1

= 77 – 27= 50°C

Heat of combustion = 4 × 104 J/g

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

Mass of flowing water, m = 3.0 litre/min = 3000 g/min

Total heat used, ΔQ = mc ΔT

= 3000 × 4.2 × 50

= 6.3 × 10J/min

∴ Rate of consumption = `(6.3 xx 0^5)/(4xx10^4) = 15.75 "g/min"`

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Solution 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 Concept: Specific Heat Capacity.
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