heat conduction

Consider an elementary annular slab at distance r fom the centre of the pipie with thickness dr, using the heat conduction equation,

Q = k.A.(dT/dr)
where Q is the rate of loss of heat
k is the thermal conductivity of glass wool
A is the area of contact
dT/dr is the temperature gradient

hence, for a length L of the pipe,
Q = k(2.pi.r.L)(dT/dr) , where pi = 3.14159 .......
Q.[dr/r] = 2.pi.k.L.(dT)
Integrating both sides, the left hand side takes the limit from 5 cm (=0.05 m) to 6 cm (=0.06 m). The right hand side takes the limit from 82'C to 15'C

Q.ln[0.06/0.05] = 2.pi.k.L.(15-82)
i.e. Q/L = -2.pi.k.(67)/ln(0.06/0.05) = - 2.pi.(0.042).(67)/ln(0.06/0.05) W/m
hence, Q/L = - 97 W/m (the -ve sign indicates heat loss)