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Thermodynamics

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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−1?4.0 \times 104 \, \mathrm{J} \, \mathrm{g}^{-1}?
Solution:  
Here, mass of water heated, m=3000 g min−1m = 3000 \, \mathrm{g} \, \mathrm{min}^{-1}
Volume of water heated =3.0=3.0 lit min −1^{-1}
Rise in temperature ΔT=(77−27)∘C=50∘C\Delta T = (77-27)^{\circ}\mathrm{C} = 50^{\circ}\mathrm{C}
Specific heat of water c=4.2 Jg−1 K−1c = 4.2 \, \mathrm{Jg}^{-1} \, \mathrm{K}^{-1}
Amount of heat used ΔQ=mcΔT\Delta Q = m c \Delta T
Now, ΔQΔt=mΔtcΔT\frac{\Delta Q}{\Delta t} = \frac{m}{\Delta t} c \Delta T
⇒ΔQDt=30001 min⁡×4.2×50;\Rightarrow \frac{\Delta Q}{D t} = \frac{3000}{1 \, \min} \times 4.2 \times 50 ;
ΔQΔt=63000 J min−1\frac{\Delta Q}{\Delta t} = 63000 \, \mathrm{J} \, \mathrm{min}^{-1}
Heat of consumption (ΔQ)=ΔmL(\Delta Q) = \Delta m L
ΔQΔt=ΔmΔt⋅L\frac{\Delta Q}{\Delta t} = \frac{\Delta m}{\Delta t} \cdot L
⇒630000 J min−1\Rightarrow 630000 \, \mathrm{J} \, \mathrm{min}^{-1}
=(ΔmΔt)×4×104 Jg−1= \left( \frac{\Delta m}{\Delta t} \right) \times 4 \times 10^{4} \, \mathrm{Jg}^{-1}
Rate of consumption of fuel
ΔmΔt=63×1044×104\frac{\Delta m}{\Delta t} = \frac{63 \times 10^{4}}{4 \times 10^{4}}
=15.75≈16 g min−1= 15.75 \approx 16 \, \mathrm{g} \, \mathrm{min}^{-1}
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