`1.56 × 10^5 J` of heat is conducted through is `2 m^2` wall of 12 cm thick in one hour. Temperature difference between the two sides of the wall is `20 ^@C.` The thermal conductivity of the material of the wall is (in `Wm ^(-1) K^(-1)`)

Options

A. 0.11
B. 0.13
C. 0.15
D. 1.2

Answer

To find the thermal conductivity of the wall material, we can use the formula for heat conduction given by Fourier's law:

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

Where:
- \( Q \) = heat conducted (in Joules)
- \( k \) = thermal conductivity (in \( W \, m^{-1} \, K^{-1} \))
- \( A \) = area of the wall (in \( m^2 \))
- \( T_1 - T_2 \) = temperature difference (in Kelvin or Celsius)
- \( L \) = thickness of the wall (in meters)
- \( t \) = time (in seconds)

Given:
- \( Q = 1.56 \times 10^5 \, J \)
- \( A = 2 \, m^2 \)
- \( L = 12 \, cm = 0.12 \, m \) (conversion from cm to m)
- \( T_1 - T_2 = 20 \, ^\circ C \) (this can be used as is)
- \( t = 1 \, hour = 3600 \, seconds \) (conversion from hours to seconds)

We need to rearrange the formula to solve for \( k \):

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

Now, substituting the values into the equation:

\[ k = \frac{(1.56 \times 10^5) \cdot (0.12)}{(2) \cdot (20) \cdot (3600)} \]

Calculating the denominator:

\[ 2 \cdot 20 \cdot 3600 = 144000 \]

Now substituting back into the equation for \( k \):

\[ k = \frac{1.56 \times 10^5 \cdot 0.12}{144000} \]

Calculating the numerator:

\[ 1.56 \times 10^5 \cdot 0.12 = 18720 \]

Now substituting into the equation for \( k \):

\[ k = \frac{18720}{144000} \]

Calculating \( k \):

\[ k = 0.13 \, W \, m^{-1} \, K^{-1} \]

Thus, the thermal conductivity of the material of the wall is:

\[ \boxed{0.13 \, W \, m^{-1} \, K^{-1}} \]