Question

The brick wall (of thermal conductivity 0.35 W/m ·◦ C) of a building has dimensions of 2.7 m by 6 m and is 16 cm thick. How much heat flows through the wall in a 18.6 h period when the average inside and outside temperatures are, respectively, 31◦C and 8◦C? Answer in units of MJ.

Answer 1

Answer:

$Q = 54.577\,MJ$

Explanation:

The heat transfer through brick wall is:

$\dot Q = \frac{k\cdot A}{L}\cdot \Delta T$

$\dot Q = \frac{\left(0.35\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (2.7\,m)\cdot (6\,m)}{0.16\,m} \cdot (31^{\circ}C - 8^{\circ}C)$

$\dot Q = 815.063\,W$

The heat flow in a 18.6-h period is:

$Q = \dot Q \cdot \Delta t$

$Q = (815.063\,W)\cdot (18.6\,h)\cdot \left(\frac{3600\,s}{1\,h} \right)$

$Q = 54576618.48\,J$

$Q = 54.577\,MJ$

Answer 2

Answer:

W = 54.6 MJ ... ( 3 sig fig )

Explanation:

Given:-

- The thermal conductivity of wall, k = 0.35 W/m°C

- The thickness of wall , L = 16 cm

- The surface dimension of wall A = ( 2.7 x 6 ) m

- The time duration t = 18.6 hours

- The inside temperature, Ti = 31°C

- The outside temperature, To = 8°C

Find:-

How much heat flows through the wall in a 18.6 h period

Solution:-

- The Fourier's law of heat conduction in ( one - dimension ) through any material with thermal conductivity "k" is represented by the rate of heat transfer in the direction of x.

                      $Q= - k*A*\frac{dT}{dx}$

- The fully derived expression for conduction heat transfer is given by:

                     $Q = k*A*\frac{T_i - T_o}{L}$

- Plug in the given values and compute the rate of heat transfer:

                    $Q = 0.35*2.7*6*\frac{31 - 8}{0.16}\\\\Q = 5.67*\frac{23}{0.16}\\\\Q = 815.0625 W$

- The heat energy that flows through the wall during time t = 18.6 hrs is given by W:

                   W = Q*t*3600 / 10^6

                   W = 815.0625*18.6*3600 / 10^6

                   W = 54576585 / 10^6 MJ

                   W = 54.6 MJ ... ( 3 sig fig )