A brass boiler has a base area of `0.15 "m"^(2)` and thickness is `1.0 "cm"`. It boils water at the rate of `6.0 "kg/min"`. When placed on a gas stove. Estimate the temperature of the part of the flame in contact. With the boiler. Thermal conductivity of brass `= 109 "Wm"^(-1) "K"^(-1)`

Explanation

To solve the problem, we will use the formula for heat conduction through a material, which is given by Fourier's law:

\[ Q = \frac{K \cdot A \cdot (T_f - T_b)}{L} \]

Where:
- \( Q \) is the heat transfer per unit time (in watts, W),
- \( K \) is the thermal conductivity of the material (in \( W/(m \cdot K) \)),
- \( A \) is the area of the base (in \( m^2 \)),
- \( T_f \) is the temperature of the flame (in °C),
- \( T_b \) is the temperature of the boiling water (100 °C),
- \( L \) is the thickness of the material (in meters).

### Step 1: Convert the thickness from cm to meters
Given thickness \( L = 1.0 \, \text{cm} = 0.01 \, \text{m} \).

### Step 2: Calculate the heat required to boil water
The mass of water boiled per minute is given as \( 6.0 \, \text{kg/min} \). We need to convert this to kg/s:
\[ \text{Mass per second} = \frac{6.0 \, \text{kg}}{60 \, \text{s}} = 0.1 \, \text{kg/s} \]

The latent heat of vaporization of water \( L_f \) is approximately \( 2256 \, \text{J/g} \) or \( 2256000 \, \text{J/kg} \).

The heat required per second (power) to boil the water is:
\[ Q = \text{mass per second} \times L_f = 0.1 \, \text{kg/s} \times 2256000 \, \text{J/kg} = 225600 \, \text{W} \]

### Step 3: Set up the equation using Fourier's law
Using the heat conduction equation:
\[ 225600 = \frac{109 \cdot 0.15 \cdot (T_f - 100)}{0.01} \]

### Step 4: Solve for \( T_f \)
Rearranging the equation:
\[ 225600 = \frac{109 \cdot 0.15 \cdot (T_f - 100)}{0.01} \]
\[ 225600 = 1635 \cdot (T_f - 100) \]
\[ T_f - 100 = \frac{225600}{1635} \]
\[ T_f - 100 = 138 \]
\[ T_f = 238 \, \text{°C} \]

### Final Answer
The estimated temperature of the part of the flame in contact with the boiler is \( T_f = 238 \, \text{°C} \).

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