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

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

Step-by-step 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*(dT)/(dx)`

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

`Q = k*A*(T_i - T_o)/(L)`

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

`Q = 0.35*2.7*6*(31 - 8)/(0.16)\\\\Q = 5.67*(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 )

Answer 2

`Q = 54.577\,MJ`

Step-by-step explanation:

The heat transfer through brick wall is:

`\dot Q = (k\cdot A)/(L)\cdot \Delta T`

`\dot Q = (\left(0.35\,(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((3600\,s)/(1\,h) \right)`

`Q = 54576618.48\,J`

`Q = 54.577\,MJ`