A brass boiler of base area `0.10 m^2` and thickness 1.0 boils water at the rate of `5.0 kg "min"^(-1)` when placed on a stove. The thermal conductivity of brass is `109 J s^(-1) m^(-1) ""^@C^(-1)` and heat of vaporisation of water is `2256 Jg^(-1)` . If the temperature of the part of the flame in contact with the boiler is `T = (3^p 2^q 5^r 11^s)/(109)` , then find the value of p+q+r+s.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
- Base area of the boiler, \( A = 0.10 \, m^2 \)
- Thickness of the boiler, \( d = 1.0 \, cm = 0.01 \, m \)
- Thermal conductivity of brass, \( k = 109 \, J \, s^{-1} \, m^{-1} \, °C^{-1} \)
- Rate of boiling water, \( \frac{dm}{dt} = 5.0 \, kg \, min^{-1} = \frac{5.0}{60} \, kg \, s^{-1} \)
- Heat of vaporization of water, \( L = 2256 \, J \, g^{-1} = 2256000 \, J \, kg^{-1} \) (since \( 1 \, kg = 1000 \, g \))

### Step 2: Calculate the Heat Required to Boil Water
The heat required to boil water in one minute is given by:
\[ Q = m \cdot L \]
Where \( m \) is the mass of water boiled in one minute.
Converting the mass from kg to grams:
\[ m = 5.0 \, kg = 5000 \, g \]
Thus, the heat required is:
\[ Q = 5000 \, g \cdot 2256 \, J/g = 11280000 \, J \]

### Step 3: Calculate the Heat Flow through the Boiler
The heat flow \( Q \) through the boiler can be expressed using Fourier's law of heat conduction:
\[ Q = \frac{k \cdot A \cdot (T - T_0) \cdot t}{d} \]
Where:
- \( T \) is the temperature of the flame,
- \( T_0 = 100 \, °C \) (boiling point of water),
- \( t = 60 \, s \) (time in seconds).

Substituting the known values:
\[ 11280000 = \frac{109 \cdot 0.10 \cdot (T - 100) \cdot 60}{0.01} \]

### Step 4: Simplify the Equation
Rearranging the equation:
\[ 11280000 = \frac{109 \cdot 0.10 \cdot 60 \cdot (T - 100)}{0.01} \]
\[ 11280000 = 65400 \cdot (T - 100) \]

### Step 5: Solve for \( T \)
Now, we can solve for \( T \):
\[ T - 100 = \frac{11280000}{65400} \]
Calculating the right-hand side:
\[ T - 100 = 172.5 \]
Thus,
\[ T = 172.5 + 100 = 272.5 \, °C \]

### Step 6: Express \( T \) in the Required Form
Now we need to express \( T \) in the form:
\[ T = \frac{3^p \cdot 2^q \cdot 5^r \cdot 11^s}{109} \]
First, we convert \( 272.5 \) to a fraction:
\[ T = 272.5 = \frac{2725}{10} = \frac{2725 \cdot 10}{10} = \frac{27250}{100} \]
Next, we need to factor \( 27250 \):
\[ 27250 = 2 \cdot 5^2 \cdot 11 \cdot 3^3 \]

### Step 7: Identify the Values of \( p, q, r, s \)
From the factorization:
- \( p = 3 \) (from \( 3^3 \))
- \( q = 1 \) (from \( 2^1 \))
- \( r = 2 \) (from \( 5^2 \))
- \( s = 1 \) (from \( 11^1 \))

### Step 8: Calculate \( p + q + r + s \)
Now we can find:
\[ p + q + r + s = 3 + 1 + 2 + 1 = 7 \]

### Final Answer
Thus, the value of \( p + q + r + s \) is \( \boxed{7} \).