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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 × 10^{4} J/g?
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Solution 1
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, T_{1} = 27°C
Final temperature, T_{2} = 77°C
∴Rise in temperature, ΔT = T_{2} – T_{1}
= 77 – 27= 50°C
Heat of combustion = 4 × 10^{4} 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 × 10^{5 }J/min
∴ Rate of consumption = `(6.3 xx 0^5)/(4xx10^4) = 15.75 "g/min"`
Solution 2
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)`
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