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?
Answers
Explanation:
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 × 105 J/min
∴ Rate of consumption = 6.3 × 105 / (4 × 104) = 15.75 g/min.
AnswEr:-
15.75 g/min.
EXplAnatIon :-
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.
The geyser heats the water, raising the temperature from 27°C to 77°C.Initial temperature, T1 = 27°C
The geyser heats the water, raising the temperature from 27°C to 77°C.Initial temperature, T1 = 27°CFinal temperature, T2 = 77°C
∴Rise in temperature, ΔT = T2 – T1
= 77 – 27 = 50°C
Heat of combustion = 4 × 104 J/g
Heat of combustion = 4 × 104 J/gSpecific heat of water, c = 4.2 J g–1 °C–1
Heat of combustion = 4 × 104 J/gSpecific heat of water, c = 4.2 J g–1 °C–1Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mc ΔT
Total heat used, ΔQ = mc ΔT= 3000 × 4.2 × 50
Total heat used, ΔQ = mc ΔT= 3000 × 4.2 × 50= 6.3 × 105 J/min
∴ Rate of consumption = 6.3 × 105 / (4 × 104)
= 15.75 g/min.