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

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

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