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1 April, 12:03

A+slab+of+a+thermal+insulator is 100cm2 in crss-section and 2cm thick. If it has a thermal conductivity of 0.1J/s/m / (C°) and a temperature difference of 100°C between opposite faces: calculate the heat flow through the slab in a day

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  1. 1 April, 12:13
    0
    The heat flow through the slab in a day is 432,000 J

    Explanation:

    Q = kA∆T/t

    Q is rate of heat flow

    k is thermal conductivity = 0.1 J/s/m.°C

    A is area = 100 cm^2 = 100 cm^2 * (1 m/100 cm) ^2 = 0.01 m^2

    ∆T is temperature difference = 100 °C

    t is thickness = 2 cm = 2/100 = 0.02 m

    Q = 0.1*0.01*100/0.02 = 5 J/s

    In a day, there are 24*60*60 = 86,400 s

    Heat flow in a day = 5 J/s * 86,400 s = 432,000 J
  2. 1 April, 12:21
    0
    Q = kA∆T/t

    Q = heat flow

    K is thermal conductivity in W/mK

    K is thermal conductivity of the material in W/mK

    ∆T is change in temperature across the material

    A is cross-sectional area in m²

    d is thickness in m

    k is thermal conductivity = 0.1 J/s/m.°C

    A is area = 100 cm^2 = 100 cm^2 * (1 m/100 cm) ^2 = 0.01 m^2

    ∆T is temperature difference = 100 °C

    t is thickness = 2 cm = 2/100 = 0.02 m

    Q = 0.1*0.01*100/0.02 = 5 J/s

    In a day, there are 24*60*60 = 86,400 s

    Heat flow in a day = 5 J/s * 86,400 s = 432,000 J
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