A 5.7 A current is set up in a circuit for 15.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced?

Answer

To find out how much the chemical energy of the battery is reduced, we can use the formula:

\[ \Delta E = V \times I \times T \]

Where:
- \(\Delta E\) is the change in chemical energy (in joules),
- \(V\) is the emf of the battery (in volts),
- \(I\) is the current (in amperes),
- \(T\) is the time (in seconds).

### Step 1: Identify the values given in the problem
- Current, \(I = 5.7 \, \text{A}\)
- Emf of the battery, \(V = 6.0 \, \text{V}\)
- Time, \(T = 15.0 \, \text{min}\)

### Step 2: Convert time from minutes to seconds
To convert minutes to seconds, we multiply by 60:
\[ T = 15.0 \, \text{min} \times 60 \, \text{s/min} = 900 \, \text{s} \]

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the formula:
\[ \Delta E = 6.0 \, \text{V} \times 5.7 \, \text{A} \times 900 \, \text{s} \]

### Step 4: Calculate the energy
Now, we perform the multiplication:
\[ \Delta E = 6.0 \times 5.7 \times 900 \]
Calculating step-by-step:
1. \(6.0 \times 5.7 = 34.2\)
2. \(34.2 \times 900 = 30780 \, \text{J}\)

### Step 5: Convert joules to kilojoules
To convert joules to kilojoules, we divide by 1000:
\[ \Delta E = \frac{30780 \, \text{J}}{1000} = 30.78 \, \text{kJ} \]

### Final Answer
The chemical energy of the battery is reduced by **30.78 kJ**.

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