A geyser heats water flowing at the rate of `3.0` liter per minute from `27^(@)C` to `77^(@)C`. If the geyser operates on a gas burner and its heat of combustion is `4.0 xx 10^(4) J//g`, then what is the rate of combustion of fuel (approx.)?

Options

A. `24 g//min`
B. `12 g//min`
C. `32 g//min`
D. `16 g//min`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the mass of water flowing per minute
Given that the geyser heats water at a rate of 3.0 liters per minute, we need to convert this volume into mass. The density of water is approximately 1 kg/L, which means:
\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 3.0 \, \text{L} \times 1000 \, \text{g/L} = 3000 \, \text{g} \]

### Step 2: Calculate the change in temperature
The water is heated from \(27^\circ C\) to \(77^\circ C\). The change in temperature (\(\Delta T\)) can be calculated as:
\[ \Delta T = 77^\circ C - 27^\circ C = 50^\circ C \]

### Step 3: Calculate the amount of heat required
The amount of heat (\(Q\)) required to heat the water can be calculated using the formula:
\[ Q = m \cdot s \cdot \Delta T \]
Where:
- \(m = 3000 \, \text{g}\) (mass of water)
- \(s = 4.2 \, \text{J/g}^\circ C\) (specific heat capacity of water)
- \(\Delta T = 50^\circ C\)

Substituting the values:
\[ Q = 3000 \, \text{g} \times 4.2 \, \text{J/g}^\circ C \times 50^\circ C = 630000 \, \text{J} \]

### Step 4: Calculate the rate of combustion of fuel
We know that the heat of combustion of the fuel is \(4.0 \times 10^4 \, \text{J/g}\). To find the rate of combustion of fuel, we need to determine how many grams of fuel are required to produce the heat calculated in Step 3.

Using the formula:
\[ \text{Rate of combustion} = \frac{Q}{\text{Heat of combustion}} = \frac{630000 \, \text{J}}{4.0 \times 10^4 \, \text{J/g}} \]

Calculating this gives:
\[ \text{Rate of combustion} = \frac{630000}{40000} = 15.75 \, \text{g/min} \]

Rounding this to the nearest whole number, we find:
\[ \text{Rate of combustion} \approx 16 \, \text{g/min} \]

### Final Answer
The rate of combustion of fuel is approximately **16 grams per minute**.

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