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 t

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.

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Minh Khuê 5 years 2021-07-25T02:29:41+00:00 2 Answers 17 views 0

Answers ( )

  1. Helga
    0
    2021-07-25T02:30:47+00:00
    Reply

    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 )

  2. Neala
    0
    2021-07-25T02:30:58+00:00
    Reply

    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

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