One end of a brass bar is maintained at `306^(@)C`, while the other end is kept at a constant, but lower, temperature. The cross-sectional area of the bar is `2.6 xx 10^(-4) m^(2)`. Because of insulation, there is negligible heat loss through the sides of the bar . Heat flows through the bar, however, at the rate 3.6 J `//` s . What is the temperature of the bar at a point 0.15m from the hot end ?

Options

A. `301^(@)C`
B. `297^(@)C`
C. `287^(@)C`
D. `265^(@)C`

Answer

To find the temperature of the brass bar at a point 0.15 m from the hot end, we can use the formula derived from Fourier's law of heat conduction:

\[ \frac{Q}{t} = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \]

Where:
- \( Q/t \) is the rate of heat transfer (in Joules per second),
- \( k \) is the thermal conductivity of the material (for brass, \( k = 110 \, \text{J/(s m °C)} \)),
- \( A \) is the cross-sectional area,
- \( T_1 \) is the temperature at the hot end,
- \( T_2 \) is the temperature at the cold end,
- \( L \) is the length of the bar.

### Step 1: Identify the known values
- \( Q/t = 3.6 \, \text{J/s} \)
- \( T_1 = 306 \, °C \)
- \( A = 2.6 \times 10^{-4} \, \text{m}^2 \)
- \( L = 0.15 \, \text{m} \)
- \( k = 110 \, \text{J/(s m °C)} \)

### Step 2: Rearranging the formula
We need to find \( T_2 \) (the temperature at the cold end). Rearranging the formula gives:

\[ T_2 = T_1 - \frac{Q/t \cdot L}{k \cdot A} \]

### Step 3: Substitute the known values into the equation
Substituting the known values into the rearranged equation:

\[ T_2 = 306 - \frac{3.6 \cdot 0.15}{110 \cdot 2.6 \times 10^{-4}} \]

### Step 4: Calculate the denominator
Calculating the denominator:

\[ 110 \cdot 2.6 \times 10^{-4} = 0.0286 \, \text{J/(s °C)} \]

### Step 5: Calculate the fraction
Now calculate the fraction:

\[ \frac{3.6 \cdot 0.15}{0.0286} = \frac{0.54}{0.0286} \approx 18.88 \]

### Step 6: Calculate \( T_2 \)
Now substitute this back into the equation for \( T_2 \):

\[ T_2 = 306 - 18.88 \approx 287.12 \, °C \]

### Final Answer
The temperature of the bar at a point 0.15 m from the hot end is approximately \( 287.12 \, °C \).