i) Thermal resistance, `R=l/(KA) = l/(K(pir^(2)))`
or `R=(2)/((401)(pi)(10^(-2)))^(2) =15.9kW^(-1)`
ii) Thermal current, `H=(DeltaT)/R = 100/15.9` or H=6.3 W
Temperature gradient
`(DeltaT)/(dx)=(0-100)/2 = -50 Km^(-1) = -50^(@)Cm^(-1)`
iv)
Let `T^(@)`C be the temperature at 25 cm from the hot end, then
T-100 = (Temperature gradient) x (distance)
or T-100 = `(-50)(0.25)`or `T=87.5^(@)`C