A very large swimming pool filled with water of temperture equal to `20^(@)`C is heated by a resistor with a heating power of 500 W for 20 minutes. Assuming the water in the pool is not in any contact with anything besides the resistor, determine the following quanties.
Is the change of entropy of the water positive, negative or zero?
`(i) DeltaS_("pool") gt 0 " " (ii) DeltaS_("pool")=0" "(iii) DeltaS_("pool") lt 0`

Answer

To determine the change in entropy of the water in the swimming pool when it is heated by a resistor, we can follow these steps:

### Step 1: Calculate the total energy supplied to the water
The power of the resistor is given as 500 W (watts), and it operates for 20 minutes. First, we need to convert the time from minutes to seconds.

\[ \text{Time in seconds} = 20 \text{ minutes} \times 60 \text{ seconds/minute} = 1200 \text{ seconds} \]

Now, we can calculate the total energy (Q) supplied to the water using the formula:

\[ Q = \text{Power} \times \text{Time} \]

\[ Q = 500 \text{ W} \times 1200 \text{ s} = 600000 \text{ J} \]

### Step 2: Calculate the change in temperature of the water
Since we are not given the specific heat capacity of water or the mass of the water in the pool, we can denote the mass of the water as \( m \) and the specific heat capacity of water as \( c \) (approximately \( 4.18 \, \text{J/g°C} \)). The change in temperature (\( \Delta T \)) can be expressed as:

\[ Q = mc\Delta T \]

From this, we can express the change in temperature:

\[ \Delta T = \frac{Q}{mc} \]

### Step 3: Determine the change in entropy
The change in entropy (\( \Delta S \)) for a system can be calculated using the formula:

\[ \Delta S = \frac{Q}{T} \]

Where \( T \) is the absolute temperature in Kelvin. The initial temperature of the water is \( 20 \, \text{°C} \), which is equivalent to:

\[ T = 20 + 273.15 = 293.15 \, \text{K} \]

Now substituting the values into the entropy change formula:

\[ \Delta S = \frac{600000 \text{ J}}{293.15 \text{ K}} \approx 2045.5 \text{ J/K} \]

### Step 4: Determine the sign of the change in entropy
Since the change in entropy \( \Delta S \) is a positive value, we conclude that the change of entropy of the water is positive.

### Final Answer
The change of entropy of the water is positive. Therefore, the correct option is:

**(i) \( \Delta S_{\text{pool}} > 0 \)**

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