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 $4.0 × 104J/g$?
✓
Answer
Water is flowing at a rate of $3.0$ litre/min. The geyser heats the water, raising the temperature from $27^\circ C$ to $77^\circ C$. Initial temperature, $T_1 = 27^\circ C$ Final temperature, $T_2 = 77^\circ C$
$\therefore$ Rise in temperature, $\triangle T = T_2 - T_1 = 77 – 27= 50^\circ C$ Heat of combustion $= 4 \times 10^4J/g$ Specific heat of water, $c = 4.2J {g^{-1}{^\circ C}^{-1}}$ Mass of flowing water, m = 3.0 litre/min = 3000g/min Total heat used, $\Delta\text{Q} =\text{mc}\Delta\text{T}$ = $3000 \times 4.2 \times 50 = 6.3 \times 10^5J/min$
$\therefore$ Rate of consumption $=\frac{6.3\times0^5}{4\times10^4}=15.75\text{g/min}$
$\therefore$ Rise in temperature, $\triangle T = T_2 - T_1 = 77 – 27= 50^\circ C$ Heat of combustion $= 4 \times 10^4J/g$ Specific heat of water, $c = 4.2J {g^{-1}{^\circ C}^{-1}}$ Mass of flowing water, m = 3.0 litre/min = 3000g/min Total heat used, $\Delta\text{Q} =\text{mc}\Delta\text{T}$ = $3000 \times 4.2 \times 50 = 6.3 \times 10^5J/min$
$\therefore$ Rate of consumption $=\frac{6.3\times0^5}{4\times10^4}=15.75\text{g/min}$
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