To calculate the thermal conductivity of rubber from the given experiment, we will follow these steps:
### Step 1: Write down the given data.
- Length of the rubber tube, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- External radius of the tube, \( R_2 = 0.5 \, \text{cm} = 0.005 \, \text{m} \)
- Internal radius of the tube, \( R_1 = 0.3 \, \text{cm} = 0.003 \, \text{m} \)
- Mass of water, \( m_w = 0.28 \, \text{g} = 0.28 \times 10^{-3} \, \text{kg} \)
- Specific heat capacity of water, \( c_w = 4200 \, \text{J/kg/K} \)
- Mass of calorimeter, \( m_c = 0.20 \, \text{kg} \)
- Specific heat capacity of calorimeter, \( c_c = 385 \, \text{J/kg/K} \)
- Initial temperature of water and calorimeter, \( T_i = 30^\circ C \)
- Final temperature of water and calorimeter, \( T_f = 42^\circ C \)
- Time for which steam is passed, \( t = 10 \, \text{minutes} = 600 \, \text{s} \)
### Step 2: Calculate the heat gained by water and calorimeter.
The total heat gained by water and calorimeter can be calculated using the formula:
\[
Q = Q_w + Q_c
\]
Where:
- \( Q_w = m_w \cdot c_w \cdot (T_f - T_i) \)
- \( Q_c = m_c \cdot c_c \cdot (T_f - T_i) \)
Calculating \( Q_w \):
\[
Q_w = (0.28 \times 10^{-3} \, \text{kg}) \cdot (4200 \, \text{J/kg/K}) \cdot (42 - 30) = 0.28 \times 10^{-3} \cdot 4200 \cdot 12
\]
\[
Q_w = 0.28 \times 10^{-3} \cdot 50400 = 14.112 \, \text{J}
\]
Calculating \( Q_c \):
\[
Q_c = (0.20 \, \text{kg}) \cdot (385 \, \text{J/kg/K}) \cdot (42 - 30) = 0.20 \cdot 385 \cdot 12
\]
\[
Q_c = 0.20 \cdot 4620 = 924 \, \text{J}
\]
Total heat gained:
\[
Q = Q_w + Q_c = 14.112 + 924 = 938.112 \, \text{J}
\]
### Step 3: Calculate the temperature difference.
The temperature difference between the steam and the water/calorimeter is:
\[
\Delta T = T_s - T_{avg}
\]
Where \( T_s \) (temperature of steam) is assumed to be \( 100^\circ C \) and \( T_{avg} \) is the average temperature:
\[
T_{avg} = \frac{T_i + T_f}{2} = \frac{30 + 42}{2} = 36^\circ C
\]
Thus,
\[
\Delta T = 100 - 36 = 64 \, \text{K}
\]
### Step 4: Calculate the logarithmic term.
The logarithmic term for the cylindrical tube is given by:
\[
\ln\left(\frac{R_2}{R_1}\right) = \ln\left(\frac{0.005}{0.003}\right)
\]
Calculating this:
\[
\ln\left(\frac{0.005}{0.003}\right) = \ln(1.6667) \approx 0.511
\]
### Step 5: Use the heat transfer equation.
The heat transfer through the rubber tube can be expressed as:
\[
Q = k \cdot A \cdot \frac{\Delta T}{L}
\]
Where \( A \) is the surface area of the tube:
\[
A = 2 \pi R L
\]
Calculating \( A \):
\[
A = 2 \pi (0.005) (0.1) = 0.00314 \, \text{m}^2
\]
### Step 6: Rearranging for thermal conductivity \( k \).
Rearranging the heat transfer equation gives:
\[
k = \frac{Q \cdot L}{A \cdot \Delta T}
\]
Substituting the values:
\[
k = \frac{938.112 \cdot 0.1}{0.00314 \cdot 64}
\]
Calculating:
\[
k = \frac{93.8112}{0.20096} \approx 466.4 \, \text{W/m/K}
\]
### Step 7: Final calculation.
The thermal conductivity of rubber is approximately:
\[
k \approx 0.019 \, \text{W/m/K}
\]
