A brass boiler has a base area of `0.15m^2` and thickness 1.0 cm it boils water at the rate of `6.0 kg//min`, When placed on a gas. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass`=109 J//s-.^@C` ) and heat of vapourization of water `=2256 J//g`.

Options

A. `350^(@)C`
B. `237^(@)C`
C. `300^(@)C`
D. `160^(@)C`

Answer

To solve the problem step by step, we will follow the outlined method in the video transcript:

### Step 1: Convert the mass flow rate of water from kg/min to g/s
Given that the rate of boiling water is 6.0 kg/min, we convert this to grams per second:
\[ \text{Mass flow rate} = 6.0 \, \text{kg/min} = 6.0 \times 10^3 \, \text{g/min} = \frac{6.0 \times 10^3 \, \text{g}}{60 \, \text{s}} = 100 \, \text{g/s} \]

**Hint:** Remember that 1 kg = 1000 g and there are 60 seconds in a minute.

### Step 2: Calculate the rate of heat supplied to the water
Using the formula for heat transfer:
\[ \text{Heat transfer rate} (Q) = m \times L \]
where \( m \) is the mass flow rate in g/s and \( L \) is the heat of vaporization of water (2256 J/g):
\[ Q = 100 \, \text{g/s} \times 2256 \, \text{J/g} = 225600 \, \text{J/s} \]

**Hint:** Ensure you use the correct units for mass and heat of vaporization.

### Step 3: Set up the heat conduction equation
We will use the formula for heat conduction:
\[ Q = K \cdot A \cdot \frac{(T_1 - T_2)}{X} \]
where:
- \( Q \) is the heat transfer rate (225600 J/s),
- \( K \) is the thermal conductivity of brass (109 J/s·m·°C),
- \( A \) is the area of the base (0.15 m²),
- \( T_1 \) is the temperature of the flame,
- \( T_2 \) is the temperature of the cold junction (assumed to be 100°C),
- \( X \) is the thickness of the boiler (1.0 cm = 0.01 m).

### Step 4: Rearrange the equation to find \( T_1 \)
Rearranging the equation gives us:
\[ T_1 - T_2 = \frac{Q \cdot X}{K \cdot A} \]
Substituting the known values:
\[ T_1 - 100 = \frac{225600 \cdot 0.01}{109 \cdot 0.15} \]

### Step 5: Calculate \( T_1 - T_2 \)
Calculating the right side:
\[ T_1 - 100 = \frac{225600 \cdot 0.01}{109 \cdot 0.15} = \frac{2256}{16.35} \approx 137.98 \]

### Step 6: Solve for \( T_1 \)
Now, substituting back to find \( T_1 \):
\[ T_1 = 137.98 + 100 = 237.98 \, °C \]

### Final Answer
The estimated temperature of the part of the flame in contact with the boiler is approximately:
\[ T_1 \approx 238 \, °C \]

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