Question

A geyser heats water flowing at the rate of 3.0 liters per minute from 27°C to 77°C. If the geyser operates on a gas burner and if its heat of combustion is \( 4.0 \times 10^4 \text{ J g}^{-1} \), the rate of combustion of the fuel per minute is:

• A. \( 15.75 \times 10^{-3} \text{ g} \)
• B. \( 15.75 \text{ g} \)
• C. \( 252 \text{ g} \)
• D. \( 252 \times 10^{-3} \text{ g} \)

Hint:

When calculating fuel consumption, ensure consistent units for energy (Joules) and mass (grams or kilograms).

Answer 1

The Correct Option is B

Step 1: Identify what heats the water.
The fuel burns and the heat it releases goes into warming the flowing water. So heat supplied by fuel equals heat gained by water each minute.
Step 2: Mass of water per minute.
The flow is $3.0$ litres per minute, and water has density $1\,\text{kg/L}$, so $m = 3\,\text{kg}$ of water passes through each minute.
Step 3: Temperature rise.
The water goes from $27^\circ\text{C}$ to $77^\circ\text{C}$, so $\Delta T = 50^\circ\text{C}$. Using $c = 4200\,\text{J kg}^{-1}\text{K}^{-1}$.
Step 4: Heat needed per minute.
\[ Q = mc\,\Delta T = 3 \times 4200 \times 50 = 6.3\times10^{5}\ \text{J}. \]
Step 5: Convert heat into fuel mass.
Each gram of fuel gives $4.0\times10^{4}\,\text{J}$, so the mass of fuel burnt is \[ m_{\text{fuel}} = \frac{Q}{\text{heat of combustion}} = \frac{6.3\times10^{5}}{4.0\times10^{4}}. \]
Step 6: Compute.
$m_{\text{fuel}} = \dfrac{63}{4} = 15.75\ \text{g}$ per minute.
\[ \boxed{15.75\ \text{g}} \]