Lake Cayuga has a volume of water estimated to be `8.2 xx 10^(12)` litres. A power station not so far above Cayuga's waters produces electricity at the rate of `1.5 xx 10^(6)` coulombs per second at an appropriate voltage. How long would it take to electrolyse the kale ?

Answer

To solve the problem of how long it would take to electrolyze the water in Lake Cayuga, we will follow these steps:

### Step 1: Convert the volume of water from litres to millilitres.
Given:
- Volume of water = \(8.2 \times 10^{12}\) litres

Since \(1 \text{ litre} = 1000 \text{ ml}\), we can convert litres to millilitres:
\[ \text{Volume in ml} = 8.2 \times 10^{12} \text{ litres} \times 1000 \text{ ml/litre} = 8.2 \times 10^{15} \text{ ml} \]

### Step 2: Calculate the mass of water.
The density of water is approximately \(1 \text{ g/ml}\), so:
\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 8.2 \times 10^{15} \text{ ml} \times 1 \text{ g/ml} = 8.2 \times 10^{15} \text{ g} \]

### Step 3: Determine the total charge required to electrolyze the water.
The electrolysis of water can be represented by the reaction:
\[ 2H_2O \rightarrow 2H_2 + O_2 \]
The molar mass of water (H2O) is approximately \(18 \text{ g/mol}\), and the number of moles of water can be calculated as:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{8.2 \times 10^{15} \text{ g}}{18 \text{ g/mol}} \approx 4.56 \times 10^{14} \text{ mol} \]

For the electrolysis of water, 2 moles of electrons are required per mole of water. Therefore, the total number of moles of electrons needed is:
\[ \text{Moles of electrons} = 2 \times \text{Number of moles of water} = 2 \times 4.56 \times 10^{14} \approx 9.12 \times 10^{14} \text{ mol} \]

Using Faraday's constant (\(F = 96500 \text{ C/mol}\)), the total charge (\(Q\)) required is:
\[ Q = \text{Moles of electrons} \times F = 9.12 \times 10^{14} \text{ mol} \times 96500 \text{ C/mol} \approx 8.80 \times 10^{19} \text{ C} \]

### Step 4: Calculate the time required for electrolysis.
Given the rate of charge production is \(1.5 \times 10^{6} \text{ C/s}\), the time (\(t\)) required can be calculated using the formula:
\[ t = \frac{Q}{\text{Current}} = \frac{8.80 \times 10^{19} \text{ C}}{1.5 \times 10^{6} \text{ C/s}} \approx 5.87 \times 10^{13} \text{ s} \]

### Step 5: Convert time from seconds to years.
To convert seconds to hours:
\[ \text{Time in hours} = \frac{5.87 \times 10^{13} \text{ s}}{3600 \text{ s/hour}} \approx 1.63 \times 10^{10} \text{ hours} \]

To convert hours to days:
\[ \text{Time in days} = \frac{1.63 \times 10^{10} \text{ hours}}{24 \text{ hours/day}} \approx 6.79 \times 10^{8} \text{ days} \]

To convert days to years:
\[ \text{Time in years} = \frac{6.79 \times 10^{8} \text{ days}}{365 \text{ days/year}} \approx 1.86 \times 10^{6} \text{ years} \]

### Final Answer:
It would take approximately \(1.86 \times 10^{6}\) years to electrolyze the water in Lake Cayuga.

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