A pitcher with `1-mm` thick porous walls contain 10kg of water. ,Water comes to its outer surface and evaporates at the rate of `0.1gs^(-1)` . The surface area of the pitcher (one side) `=200cm^(2)` . The room temperature `=42^(@)C` , latent heat of vaporization `=2.27xx10^(6)Jkg^(-1)` , and the thermal conductivity of the porous walls `=0.80js^(-1)m^(-1)C^(-1)` . Calculate the temperature of water in the pitcher when it attains a constant value.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert Units
First, we need to convert the given units into SI units for consistency.

- The thickness of the wall is given as `1 mm`, which is `1 x 10^(-3) m`.
- The surface area of the pitcher is given as `200 cm²`, which is `200 x 10^(-4) m²` or `0.02 m²`.
- The rate of evaporation is given as `0.1 g/s`, which is `0.1 x 10^(-3) kg/s` or `0.0001 kg/s`.

### Step 2: Write the Heat Conduction Equation
The heat conducted through the wall can be calculated using the formula:

\[ Q = \frac{k \cdot A \cdot (T_{room} - T_w)}{d} \]

Where:
- \( Q \) = heat conducted (J/s)
- \( k \) = thermal conductivity of the wall = `0.80 J/(s·m·°C)`
- \( A \) = surface area = `0.02 m²`
- \( T_{room} \) = room temperature = `42 °C`
- \( T_w \) = temperature of the water (unknown)
- \( d \) = thickness of the wall = `1 x 10^(-3) m`

### Step 3: Write the Heat Required for Evaporation
The heat required for the evaporation of water can be calculated using the formula:

\[ Q_{evap} = \dot{m} \cdot L \]

Where:
- \( \dot{m} \) = mass flow rate = `0.0001 kg/s`
- \( L \) = latent heat of vaporization = `2.27 x 10^6 J/kg`

### Step 4: Set the Heat Conducted Equal to the Heat Required for Evaporation
At steady state, the heat conducted through the wall equals the heat required for evaporation:

\[ \frac{k \cdot A \cdot (T_{room} - T_w)}{d} = \dot{m} \cdot L \]

### Step 5: Substitute the Known Values
Substituting the values into the equation:

\[ \frac{0.80 \cdot 0.02 \cdot (42 - T_w)}{1 \times 10^{-3}} = 0.0001 \cdot (2.27 \times 10^6) \]

### Step 6: Simplify the Equation
Calculating the left side:

\[ \frac{0.80 \cdot 0.02}{1 \times 10^{-3}} = 16 \]

Thus, the equation becomes:

\[ 16 \cdot (42 - T_w) = 0.0001 \cdot 2.27 \times 10^6 \]

Calculating the right side:

\[ 0.0001 \cdot 2.27 \times 10^6 = 227 \]

So we have:

\[ 16 \cdot (42 - T_w) = 227 \]

### Step 7: Solve for \( T_w \)
Dividing both sides by 16:

\[ 42 - T_w = \frac{227}{16} \]

Calculating the right side:

\[ \frac{227}{16} \approx 14.1875 \]

Thus, we have:

\[ 42 - T_w = 14.1875 \]

Rearranging gives:

\[ T_w = 42 - 14.1875 \approx 27.8125 \approx 28 °C \]

### Final Answer
The temperature of the water in the pitcher when it attains a constant value is approximately **28 °C**.

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